1 chapter 4: probability. 2 experiment, outcomes, and sample space simple and compound events
TRANSCRIPT
1
CHAPTER 4
PROBABILITY
2
EXPERIMENT OUTCOMES AND SAMPLE SPACE
Simple and Compound Events
3
EXPERIMENT OUTCOMES AND SAMPLE SPACE
Definition An experiment is a process that
when performed results in one and only one of many observations These observations are called that outcomes of the experiment The collection of all outcomes for an experiment is called a sample space
4
Table 41 Examples of Experiments Outcomes and Sample Spaces
Experiment Outcomes Sample SpaceToss a coin onceRoll a die onceToss a coin twicePlay lotteryTake a testSelect a student
Head Tail1 2 3 4 5 6HH HT TH TTWin LosePass FailMale Female
S = Head TailS = 1 2 3 4 5 6S = HH HT TH TTS = Win LoseS = Pass FailS = Male Female
5
Example 4-1
Draw the Venn and tree diagrams for the experiment of tossing a coin once
6
Figure 41 (a) Venn Diagram and (b) tree diagram for one toss of a coin
H
T
H
T
S Head
Tail
Outcomes
(a)(b)
7
Example 4-2
Draw the Venn and tree diagrams for the experiment of tossing a coin twice
8
Figure 42 a Venn diagram for two tosses of a coin
HH
TT
S
(a)
TH
HT
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
2
EXPERIMENT OUTCOMES AND SAMPLE SPACE
Simple and Compound Events
3
EXPERIMENT OUTCOMES AND SAMPLE SPACE
Definition An experiment is a process that
when performed results in one and only one of many observations These observations are called that outcomes of the experiment The collection of all outcomes for an experiment is called a sample space
4
Table 41 Examples of Experiments Outcomes and Sample Spaces
Experiment Outcomes Sample SpaceToss a coin onceRoll a die onceToss a coin twicePlay lotteryTake a testSelect a student
Head Tail1 2 3 4 5 6HH HT TH TTWin LosePass FailMale Female
S = Head TailS = 1 2 3 4 5 6S = HH HT TH TTS = Win LoseS = Pass FailS = Male Female
5
Example 4-1
Draw the Venn and tree diagrams for the experiment of tossing a coin once
6
Figure 41 (a) Venn Diagram and (b) tree diagram for one toss of a coin
H
T
H
T
S Head
Tail
Outcomes
(a)(b)
7
Example 4-2
Draw the Venn and tree diagrams for the experiment of tossing a coin twice
8
Figure 42 a Venn diagram for two tosses of a coin
HH
TT
S
(a)
TH
HT
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
3
EXPERIMENT OUTCOMES AND SAMPLE SPACE
Definition An experiment is a process that
when performed results in one and only one of many observations These observations are called that outcomes of the experiment The collection of all outcomes for an experiment is called a sample space
4
Table 41 Examples of Experiments Outcomes and Sample Spaces
Experiment Outcomes Sample SpaceToss a coin onceRoll a die onceToss a coin twicePlay lotteryTake a testSelect a student
Head Tail1 2 3 4 5 6HH HT TH TTWin LosePass FailMale Female
S = Head TailS = 1 2 3 4 5 6S = HH HT TH TTS = Win LoseS = Pass FailS = Male Female
5
Example 4-1
Draw the Venn and tree diagrams for the experiment of tossing a coin once
6
Figure 41 (a) Venn Diagram and (b) tree diagram for one toss of a coin
H
T
H
T
S Head
Tail
Outcomes
(a)(b)
7
Example 4-2
Draw the Venn and tree diagrams for the experiment of tossing a coin twice
8
Figure 42 a Venn diagram for two tosses of a coin
HH
TT
S
(a)
TH
HT
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
4
Table 41 Examples of Experiments Outcomes and Sample Spaces
Experiment Outcomes Sample SpaceToss a coin onceRoll a die onceToss a coin twicePlay lotteryTake a testSelect a student
Head Tail1 2 3 4 5 6HH HT TH TTWin LosePass FailMale Female
S = Head TailS = 1 2 3 4 5 6S = HH HT TH TTS = Win LoseS = Pass FailS = Male Female
5
Example 4-1
Draw the Venn and tree diagrams for the experiment of tossing a coin once
6
Figure 41 (a) Venn Diagram and (b) tree diagram for one toss of a coin
H
T
H
T
S Head
Tail
Outcomes
(a)(b)
7
Example 4-2
Draw the Venn and tree diagrams for the experiment of tossing a coin twice
8
Figure 42 a Venn diagram for two tosses of a coin
HH
TT
S
(a)
TH
HT
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
5
Example 4-1
Draw the Venn and tree diagrams for the experiment of tossing a coin once
6
Figure 41 (a) Venn Diagram and (b) tree diagram for one toss of a coin
H
T
H
T
S Head
Tail
Outcomes
(a)(b)
7
Example 4-2
Draw the Venn and tree diagrams for the experiment of tossing a coin twice
8
Figure 42 a Venn diagram for two tosses of a coin
HH
TT
S
(a)
TH
HT
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
6
Figure 41 (a) Venn Diagram and (b) tree diagram for one toss of a coin
H
T
H
T
S Head
Tail
Outcomes
(a)(b)
7
Example 4-2
Draw the Venn and tree diagrams for the experiment of tossing a coin twice
8
Figure 42 a Venn diagram for two tosses of a coin
HH
TT
S
(a)
TH
HT
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
7
Example 4-2
Draw the Venn and tree diagrams for the experiment of tossing a coin twice
8
Figure 42 a Venn diagram for two tosses of a coin
HH
TT
S
(a)
TH
HT
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
8
Figure 42 a Venn diagram for two tosses of a coin
HH
TT
S
(a)
TH
HT
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
9
Figure 42 b Tree diagram for two tosses of coin
T
H
Finaloutcomes
(b)
T TT
HHH
THHT
H
T
Secondtoss
First toss
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
10
Example 4-3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman Write all the outcomes for this experiment Draw the Venn and tree diagrams for this experiment
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
11
Figure 43 a Venn diagram for selecting two persons
MM
WW
MW
WM
S
(a)
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
12
Figure 43 b Tree diagram for selecting two persons
W
M
Finaloutcomes
(b)
W WW
MMM
WMMW
M
W
Secondselection
Firstselection
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
13
Simple and Compound Events
Definition An event is a collection of one or
more of the outcomes of an experiment
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
14
Simple and Compound Events cont
Definition An event that includes one and only
one of the (final) outcomes for an experiment is called a simple event and is denoted by Ei
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
15
Example 4-4 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Each of the final four outcomes (MM MW WM WW) for this experiment is a simple event These four events can be denoted by E1 E2 E3 and E4 respectively Thus
E1 = (MM ) E2 = (MW ) E3 = (WM ) and E4
= (WW )
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
16
Simple and Compound Events
Definition A compound event is a collection of
more than one outcome for an experiment
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
17
Example 4-5 Reconsider Example 4-3 on selecting two persons
from the members of a club and observing whether the person selected each time is a man or a woman Let A be the event that at most one man is selected Event A will occur if either no man or one man is selected Hence the event A is given by
A = MW WM WW
Because event A contains more than one outcome it is a compound event The Venn diagram in Figure 44 gives a graphic presentation of compound event A
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
18
Figure 44 Venn diagram for event A
MM
S
MW
WM WW
A
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
19
Example 4-6 In a group of a people some are in favor of genetic
engineering and others are against it Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering How many distinct outcomes are possible Draw a Venn diagram and a tree diagram for this experiment List all the outcomes included in each of the following events and mention whether they are simple or compound events
(a) Both persons are in favor of the genetic engineering(b) At most one person is against genetic engineering(c) Exactly one person is in favor of genetic engineering
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
20
Solution 4-6 Let
F = a person is in favor of genetic engineering A = a person is against genetic engineering FF = both persons are in favor of genetic
engineering FA = the first person is in favor and the second is
against AF = the first is against and the second is in favor AA = both persons are against genetic engineering
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
21
Figure 45 a Venn diagram
FF
S
FA
AF AA
(a)
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
22
Figure 45 b Tree diagram
A
F
Finaloutcomes
(b)
A AA
FFF
AFFA
F
A
Secondperson
Firstperson
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
23
Solution 4-6
a) Both persons are in favor of genetic engineering = FF
It is a simple eventb) At most one person is against genetic
engineering = FF FA AF It is a compound eventc) Exactly one person is in favor of genetic
engineering = FA AF It is a compound event
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
24
CALCULATING PROBABILITY
Two Properties of probability Three Conceptual Approaches to
Probability Classical Probability Relative Frequency Concept of
Probability Subjective Probability
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
25
CALCULATING PROBABLITY
Definition Probability is a numerical measure
of the likelihood that a specific event will occur
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
26
Two Properties of Probability
First Property of Probability 0 le P (Ei) le 1 0 le P (A) le 1
Second Property of Probability ΣP (Ei) = P (E1) + P (E2) + P (E3) + hellip = 1
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
27
Three Conceptual Approaches to Probability
Classical Probability
Definition Two or more outcomes (or events)
that have the same probability of occurrence are said to be equally likely outcomes (or events)
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
28
Classical Probability
Classical Probability Rule to Find Probability
experiment for the outcomes ofnumber Total
tofavorable outcomes ofNumber )(
experiment for the outcomes ofnumber Total
1)(
AAP
EP i
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
29
Example 4-7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
30
Solution 4-7
502
1
outcomes ofnumber Total
1)head( P
502
1tail)( P
Similarly
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
31
Example 4-8
Find the probability of obtaining an even number in one roll of a die
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
32
Solution 4-8
506
3
outcomes ofnumber Total
in included outcomes ofNumber )head(
AP
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
33
Example 4-9
In a group of 500 women 80 have played golf at lest once Suppose one of these 500 women is randomly selected What is the probability that she has played golf at least once
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
34
Solution 4-9
16500
80once)least at golf played has woman selected( P
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
35
Three Conceptual Approaches to Probability cont
Relative Concept of Probability
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times then according to the relative frequency concept of probability
n
fAP )(
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
36
Example 4-10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons Assuming that the lemons are manufactured randomly what is the probability that the next car manufactured at this auto factory is a lemon
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
37
Solution 4-10 Let n denotes the total number of cars in
the sample and f the number of lemons in n Then
n = 500 and f = 10 Using the relative frequency concept of
probability we obtain02
500
10lemon) a iscar next (
n
fP
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
38
Table 42 Frequency and Relative Frequency Distributions for the Sample of Cars
Car fRelative
frequency
GoodLemon
49010
490500 = 9810500 = 02
n = 500
Sum = 100
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
39
Law of Large Numbers
Definition Law of Large Numbers If an
experiment is repeated again and again the probability of an event obtained from the relative frequency approaches the actual or theoretical probability
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
40
Three Conceptual Approaches to Probability
Subjective Probability
Definition Subjective probability is the
probability assigned to an event based on subjective judgment experience information and belief
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
41
COUNTING RULE
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes the second step in n outcomes and the third in k outcomes then
Total outcomes for the experiment = m n k
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
42
Example 4-12 Suppose we toss a coin three times This
experiment has three steps the first toss the second toss and the third toss Each step has two outcomes a head and a tail Thus
Total outcomes for three tosses of a coin = 2 x 2 x 2 = 8
The eight outcomes for this experiment are HHH HHT HTH HTT THH THT TTH and TTT
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
43
Example 4-13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months 48 months or 60 months How many total outcomes are possible
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
44
Solution 4-13
Total outcomes = 2 x 3 = 6
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
45
Example 4-14
A National Football League team will play 16 games during a regular season Each game can result in one of three outcomes a win a lose or a tie The total possible outcomes for 16 games are calculated as followsTotal outcomes = 333333333333 3333
= 316 = 43046721 One of the 43046721 possible outcomes is
all 16 wins
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
46
MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of US companies Table 43 gives a two way classification of the responses of these 100 employees
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
47
Table 43 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
48
MARGINAL AND CONDITIONAL PROBABILITIES
Table 44 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female
4 36 40
Total 19 81 100
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
49
MARGINAL AND CONDITIONAL PROBABILITIES
Definition Marginal probability is the
probability of a single event without consideration of any other event Marginal probability is also called simple probability
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
50
Table 45 Listing the Marginal Probabilities
In Favor
(A )
Against
(B )
Total
Male (M ) 15 45 60
Female (F )
4 36 40
Total 19 81 100
P (M ) = 60100 = 60P (F ) = 40100 = 40
P (A ) = 19100
= 19
P (B ) = 81100
= 81
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
51
MARGINAL AND CONDITIONAL PROBABILITIES cont
P ( in favor | male)
The event whose probability is to be determined
This event has already occurred
Read as ldquogivenrdquo
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
52
MARGINAL AND CONDITIONAL PROBABILITIES cont
Definition Conditional probability is the probability
that an event will occur given that another has already occurred If A and B are two events then the conditional probability A given B is written as
P ( A | B ) and read as ldquothe probability of A given that
B has already occurredrdquo
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
53
Example 4-15
Compute the conditional probability P ( in favor | male) for the data on 100 employees given in Table 44
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
54
Solution 4-15
In Favor Against Total
Male 15 45 60
Males who are in favor
Total number of males
2560
15
males ofnumber Total
favorin are whomales ofNumber male) |favor in ( P
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
55
Figure 46Tree Diagram
Female
Male
Favors | Male
60100
40100
1560
4560
440
3640
Against | Male
Favors | Female
Against | Female
This event has already
occurred
We are to find the probability of this
event
Required probability
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
56
Example 4-16
For the data of Table 44 calculate the conditional probability that a randomly selected employee is a female given that this employee is in favor of paying high salaries to CEOs
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
57
Solution 4-16
In Favor
15
4
19
Females who are in favor
Total number of employees who are in favor
210519
4
favorin are whoemployees ofnumber Total
favorin are whofemales ofNumber favor)in | female(
P
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
58
Figure 47 Tree diagram
Against
Favors
Female | Favors
19100
81100
1560
419
4581
3681
Male | Favors
Male | Against
Female | Against
We are to find the probability of this
event
Required probability
This event has already
occurred
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
59
MUTUALLY EXCLUSIVE EVENTS
Definition Events that cannot occur together are
said to be mutually exclusive events
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
60
Example 4-17
Consider the following events for one roll of a die A= an even number is observed= 2 4 6 B= an odd number is observed= 1 3 5 C= a number less than 5 is observed= 1 2
3 4 Are events A and B mutually exclusive
Are events A and C mutually exclusive
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
61
Solution 4-17
Figure 48 Mutually exclusive events A and B
S
1
3
52
4
6
A
B
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
62
Solution 4-17
Figure 49 Mutually nonexclusive events A and C
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
63
Example 4-18
Consider the following two events for a randomly selected adult Y = this adult has shopped on the Internet at
least once N = this adult has never shopped on the
Internet Are events Y and N mutually exclusive
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
64
Solution 4-18
Figure 410 Mutually exclusive events Y and N
SNY
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
65
INDEPENDENT VERSUS DEPENDENT EVENTS
Definition Two events are said to be independent
if the occurrence of one does not affect the probability of the occurrence of the other In other words A and B are independent events if
either P (A | B ) = P (A ) or P (B | A ) = P (B )
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
66
Example 4-19
Refer to the information on 100 employees given in Table 44 Are events ldquofemale (F )rdquo and ldquoin favor (A )rdquo independent
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
67
Solution 4-19 Events F and A will be independent if P (F ) = P (F | A )
Otherwise they will be dependent
From the information given in Table 44P (F ) = 40100 = 40P (F | A ) = 419 = 2105
Because these two probabilities are not equal the two events are dependent
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
68
Example 4-20 A box contains a total of 100 CDs that were
manufactured on two machines Of them 60 were manufactured on Machine I Of the total CDs 15 are defective Of the 60 CDs that were manufactured on Machine I 9 are defective Let D be the event that a randomly selected CD is defective and let A be the event that a randomly selected CD was manufactured on Machine I Are events D and A independent
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
69
Solution 4-20
From the given information P (D ) = 15100 = 15
P (D | A ) = 960 = 15Hence
P (D ) = P (D | A ) Consequently the two events are
independent
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
70
Table 46 Two-Way Classification Table
Defective (D )
Good (G ) Total
Machine I (A )Machine II (B )
9 6
5134
60 40
Total 15 85 100
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
71
Two Important Observations
Two events are either mutually exclusive or independent Mutually exclusive events are always
dependent Independent events are never mutually
exclusive Dependents events may or may not
be mutually exclusive
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
72
COMPLEMENTARY EVENTS
Definition The complement of event A denoted
by Ā and is read as ldquoA barrdquo or ldquoA complementrdquo is the event that includes all the outcomes for an experiment that are not in A
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
73
Figure 411 Venn diagram of two complementary events
S
AA
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
74
Example 4-21
In a group of 2000 taxpayers 400 have been audited by the IRS at least once If one taxpayer is randomly selected from this group what are the two complementary events for this experiment and what are their probabilities
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
75
Solution The complementary events for this
experiment are A = the selected taxpayer has been audited by
the IRS at least once Ā = the selected taxpayer has never been
audited by the IRS
The probabilities of the complementary events are
P (A) = 4002000 = 20
P (Ā) = 16002000 = 80
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
76
Figure 412 Venn diagram
S
AA
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
77
Example 4-22
In a group of 5000 adults 3500 are in favor of stricter gun control laws 1200 are against such laws and 300 have no opinion One adult is randomly selected from this group Let A be the event that this adult is in favor of stricter gun control laws What is the complementary event of A What are the probabilities of the two events
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
78
Solution 4-22 The two complementary events are
A = the selected adult is in favor of stricter gun control laws
Ā = the selected adult either is against such laws or has no opinion
The probabilities of the complementary events are
P (A) = 35005000 = 70
P (Ā) = 15005000 = 30
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
79
Figure 413 Venn diagram
S
AA
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
80
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Intersection of Events Multiplication Rule
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
81
Intersection of Events
Definition Let A and B be two events defined in a
sample space The intersection of A and B represents the collection of all outcomes that are common to both A and B and is denoted by
A and B
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
82
Figure 414 Intersection of events A and B
A and B
A B
Intersection of A and B
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
83
Multiplication Rule
Definition The probability of the intersection of
two events is called their joint probability It is written as
P (A and B ) or P (A cap B )
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
84
INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE
Multiplication Rule to Find Joint Probability
The probability of the intersection of two events A and B is
P (A and B ) = P (A )P (B |A )
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
85
Example 4-23
Table 47 gives the classification of all employees of a company given by gender and college degree
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
86
Table 47 Classification of Employees by Gender and Education
College Graduate
(G )
Not a College Graduate
(N ) Total
Male (M )Female (F )
74
209
2713
Total 11 29 40
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
87
Example 4-23
If one of these employees is selected at random for membership on the employee management committee what is the probability that this employee is a female and a college graduate
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
88
Solution 4-23
Calculate the intersection of event F and G
P (F and G ) = P (F )P (G |F ) P (F ) = 1340 P (G |F ) = 413 P (F and G ) = (1340)(413) = 100
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
89
Figure 415 Intersection of events F and G
4
Females College graduates
Females and college graduates
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
90
Figure 416 Tree diagram for joint probabilities
M
F
G | M
N | M
G | F
N | F
413
913
2027
727
2440
1340
Graduates nongraduatesMale female Final outcomes
P(M and G) = (2740) (2027) = 175
P(M and N) = (2740) (2027) = 500
P(F and N) = (1340) (913) = 225
P(F and G) = (1340) (413) = 100
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
91
Example 4-24
A box contains 20 DVDs 4 of which are defective If 2 DVDs are selected at random (without replacement) from this box what is the probability that both are defective
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
92
Solution 4-24 Let us define the following events for this
experimentG1 = event that the first DVD selected is goodD1 = event that the first DVD selected is defectiveG2 = event that the second DVD selected is goodD2 = event that the second DVD selected is defective
The probability to be calculated is P (D1 and D2) = P (D1 )P (D2 |D1 ) P (D1) = 420P (D2 |D1) = 319P (D1 and D2) = (420)(319) = 0316
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
93
Figure 417 Selecting two DVDs
G1
D1
G2 | G1
D2 | G1
G2 | D1
D2 | D1
1619
319
419
1519
1620
420
Second selectionFirst selection Final outcomes
P(G1 and G2) = (1620) (1519) = 6316
P(G1 and D2) = (1620) (419) = 1684
P(D1 and G2) = (420) (1619) = 1684
P(D1 and D2) = (420) (319) = 0316
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
94
Multiplication Rule cont Calculating Conditional Probability
If A and B are two events then
given that P (A ) ne 0 and P (B ) ne 0
)(
) and ()|( and
)(
) and ()|(
BP
BAPBAP
AP
BAPABP
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
95
Example 4-25
The probability that a randomly selected student from a college is a senior is 20 and the joint probability that the student is a computer science major and a senior is 03 Find the conditional probability that a student selected at random is a computer science major given that heshe is a senior
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
96
Solution 4-25 Let us define the following two events
A = the student selected is a senior B = the student selected is a computer science
major From the given information
P (A) = 20 and P (A and B) = 03 Hence
P (B | A ) = 0320 = 15
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
97
Multiplication Rule for Independent Events
Multiplication Rule to Calculate the Probability of Independent Events
The probability of the intersection of two independent events A and B is
P (A and B ) = P (A )P (B )
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
98
Example 4-26
An office building has two fire detectors The probability is 02 that any fire detector of this type will fail to go off during a fire Find the probability that both of these fire detectors will fail to go off in case of a fire
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
99
Solution 4-26
Let A = the first fire detector fails to go off during
a fireB = the second fire detector fails to go off
during a fire
Then the joint probability of A and B isP (A and B ) = P (A) P (B ) = (02)(02)
= 0004
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
100
Example 4-27 The probability that a patient is allergic
to penicillin is 20 Suppose this drug is administered to three patients
a) Find the probability that all three of them are allergic to it
b) Find the probability that at least one of the them is not allergic to it
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
101
Solution
a) Let A B and C denote the events the first second and third patients respectively are allergic to penicillin Hence
P (A and B and C ) = P (A ) P (B ) P (C ) = (20) (20) (20) = 008
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
102
Solution
b) Let us define the following eventsG = all three patients are allergicH = at least one patient is not allergic
P (G ) = P (A and B and C ) = 008 Therefore using the complementary
event rule we obtain P (H ) = 1 ndash P (G )
= 1 - 008 = 992
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
103
Figure 418 Tree diagram for joint probabilities
A
A
B
C
B
B
B
C
C
C
C
C
C
C
20
80
80
80
80
80
80
80
20
20
20
20
20
20
P(ABC) = 008
P(ABC) = 032
P(ABC) = 032
P(ABC) = 128
P(ABC) = 032
P(ABC) = 128
P(ABC) = 512
P(ABC) = 128
First patient Second patient Third patient Final outcomes
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
104
Multiplication Rule for Independent Events
Joint Probability of Mutually Exclusive Events
The joint probability of two mutually exclusive events is always zero If A and B are two mutually exclusive events then
P (A and B ) = 0
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
105
Example 4-28
Consider the following two events for an application filed by a person to obtain a car loan A = event that the loan application is
approved R = event that the loan application is
rejected What is the joint probability of A and R
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
106
Solution 4-28
The two events A and R are mutually exclusive Either the loan application will be approved or it will be rejected Hence
P (A and R ) = 0
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
107
UNION OF EVENTS AND THE ADDITION RULE
Definition Let A and B be two events defined in a
sample space The union of events A and B is the collection of all outcomes that belong to either A or B or to both A and B and is denoted by
A or B
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
108
Example 4-29
A senior citizen center has 300 members Of them 140 are male 210 take at least one medicine on a permanent basis and 95 are male and take at least one medicine on a permanent basis Describe the union of the events ldquomalerdquo and ldquotake at least one medicine on a permanent basisrdquo
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
109
Solution 4-29 Let us define the following events
M = a senior citizen is a maleF = a senior citizen is a femaleA = a senior citizen takes at least one medicineB = a senior citizen does not take any medicine
The union of the events ldquomalerdquo and ldquotake at least one medicinerdquo includes those senior citizens who are either male or take at least one medicine or both The number of such senior citizen is
140 + 210 ndash 95 = 255
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
110
Table 48
A B Total
M 95 45 140
F 115 45 160
Total 210 90 300
Counted twice
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
111
Figure 419 Union of events M and A
M A
Shaded area gives the union of events M and A and includes 255 senior citizen
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
112
Multiplication Rule for Independent Events
Addition Rule
Addition Rule to Find the Probability of Union of Events
The portability of the union of two events A and B is P (A or B ) = P (A ) + P (B) ndash P (A and B )
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
113
Example 4-30 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue Table 49 gives a two-way classification of the responses of these faculty members and students
Find the probability that one person selected at random from these 300 persons is a faculty member or is in favor of this proposal
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
114
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
115
Solution 4-30
Let us define the following eventsA = the person selected is a faculty memberB = the person selected is in favor of the proposal
From the information in the Table 49P (A ) = 70300 = 2333P (B ) = 135300 = 4500P (A and B) = P (A) P (B | A ) = (70300)(4570)
= 1500
Using the addition rule we haveP (A or B ) = P (A ) + P (B ) ndash P (A and B ) = 2333 + 4500 ndash 1500 = 5333
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
116
Example 4-31
A total of 2500 persons 1400 are female 600 are vegetarian and 400 are female and vegetarian What is the probability that a randomly selected person from this group is a male or vegetarian
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
117
Solution 4-31 Let us define the following events
F = the randomly selected person is a female M = the randomly selected person is a male V = the randomly selected person is a vegetarian N = the randomly selected person is a non-
vegetarian
60082444 2500
200
2500
600
2500
1100
) and ()()()or (
VMPVPMPVMP
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
118
Table 410 Two-Way Classification Table
Vegetarian (V)
Nonvegetarian (N) Total
Female (F)Male (M)
400200
1000900
14001100
Total 600 1900 2500
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
119
Addition Rule for Mutually Exclusive Events
Addition Rule to Find the Probability of the Union of Mutually Exclusive Events
The probability of the union of two mutually exclusive events A and B is
P (A or B ) = P (A ) + P (B )
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
120
Example 4-32 A university president has proposed that all
students must take a course in ethics as a requirement for graduation Three hundred faculty members and students from this university were asked about their opinion on this issue The following table reproduced from Table 49 in Example 4-30 gives a two-way classification of the responses of these faculty members and students
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
121
Table 49 Two-Way Classification of Responses
Favor Oppose Neutral Total
FacultyStudent
4590
15110
1030
70230
Total 135 125 40 300
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
122
Example 4-32
What is the probability that a randomly selected person from these 300 faculty members and students is in favor of the proposal or is neutral
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
123
Figure 420 Venn diagram of mutually exclusive events
FN
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
124
Solution 4-32
Let us define the following eventsF = the person selected is in favor of the proposalN = the person selected is neutral
From the given informationP (F ) = 135300 = 4500P (N ) = 40300 = 1333
HenceP (F or N ) = P (F ) + P (N ) = 4500 + 1333
= 5833
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
125
Example 4-33
Consider the experiment of rolling a die twice Find the probability that the sum of the numbers obtained on two rolls is 5 7 or 10
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
126
Table 411 Two Rolls of a Die
Second Roll of the Die
1 2 3 4 5 6
FirstRoll ofthe Die
1 (11) (12) (13) (14) (15) (16)
2 (21) (22) (23) (24) (25) (26)
3 (31) (32) (33) (34) (35) (36)
4 (41) (42) (43) (44) (45) (46)
5 (51) (52) (53) (54) (55) (56)
6 (61) (62) (63) (64) (65) (66)
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
127
Solution 4-33
P (sum is 5 or 7 or 10) = P (sum is 5) + P (sum is 7) + P (sum is 10)= 436 + 636 + 336 = 1336 = 3611
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
128
Example 4-34
The probability that a person is in favor of genetic engineering is 55 and that a person is against it is 45 Two persons are randomly selected and it is observed whether they favor or oppose genetic engineering
a) Draw a tree diagram for this experimentb) Find the probability that at least one of the
two persons favors genetic engineering
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
129
Solution 4-34
a) Let F = a person is in favor of genetic
engineering A = a person is against genetic engineering
The tree diagram in Figure 421 shows these four outcomes and their probabilities
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
130
Figure 421 Tree diagram
First person Second person Final outcomes and theirprobabilities
F
A
F
F
A
A
55
45
55
55
55
45
45
P(FF) = (55) (55) = 3025
P(FA) = (55) (45) = 2475
P(AF) = (45) (55) = 2475
P(AA) = (45) (45) = 2025
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975
131
Solution
b) P ( at least one person favors) = P (FF or FA or AF ) = P (FF ) + P (FA ) + P (AF ) = 3025 + 2475 + 2475 = 7975