probability - nkd group · 2013. 1. 29. · probability of an event is always between 0 & 1. 2....
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Probability
The Science Of Uncertainty
BASIC PROPERTIES OF PROBABILITIES
1. Probability of an event is always between 0 & 1.
2. An impossible event has a probability of 0.
3. A guaranteed (certain) event has a probability of 1.
Chapter 4: Probability Concepts Page -2- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
SOME DEFINITIONS
1. Classical Probability
All possible outcomes of the experiment are equally likely.
That is, suppose an experiment has N possible outcomes, all
equally likely then the probability that a specified event
occurs is equal to the number of ways, f, that the event can
occur, divided by the total number of possible outcomes.
Notation: f / N
2. Outcome
The results of a trial.
3. Sample Space, or Set (S)
A collection of all possible outcomes for an experiment.
4. Event
A collection of outcomes for the experiment. That is, any
subset of the sample space. Events are generally denoted
as: A, B, C, etc.
5. Probability is a generalization of the concept of percentages.
For example if we compute the probability of getting a
double when two balanced dice are rolled as 0.167 then we
can say there is a 16.7% chance of getting a double when
two balanced dice are rolled.
Chapter 4: Probability Concepts Page -3- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Example:
An experiment of selecting a card from a deck of 52 cards was
performed. The following are the some sample events and their
outcomes.
Sample Events Outcomes
A. A card selected is a king of diamonds 1
B. A card selected is a king 4
C. A card selected is a diamond 13
D. A card selected is not a face card 40
(52-12)
Chapter 4: Probability Concepts Page -4- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
OPERATIONS WITH SETS
1. Subset --- A B
A set whose members belong to another larger set.
2. Union --- (A B) or (A or B)
The addition of two or more sets.
3. Intersection --- (A B) or (A & B)
The common elements of two or more sets.
4. Null
An empty set.
5. Complement --- (A
_
) or (not A)
All elements that do not belong to a set.
6. Mutually Exclusive (Disjoint)
When two sets have no common elements they are
considered to be disjoint.
Chapter 4: Probability Concepts Page -5- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Example For Set Operation
Consider an experiment to determine if people wear seatbelts.
Two possible outcomes: Y or N.
If three people are selected at random what is the sample space?
S = {YYY, YYN, YNY, NYY, YNN, NYN, NNY, NNN}
#of outcomes = 23 = 2 x 2 x 2 = 8
Define the following events:
A = At least two wear seatbelts
= {YYY, YYN, YNY, NYY}
B = Exactly two wear seatbelts
= {YYN, YNY, NYY}
C = Exactly one wears a seatbelt
= {YNN, NYN, NNY}
AB = {YYY, YYN, YNY, NYY}
A&B = {YYN, YNY, NYY}
A&C = Empty Set
not A = {YNN, NYN, NNY, NNN}
= at most one wears seatbelt
Chapter 4: Probability Concepts Page -6- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
A
B
A Not B A B
B Not A
Not (A or B)
Chapter 4: Probability Concepts Page -7- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
NOTATION FOR PROBABILITY
IF: E is an event
P is the probability
P(E) is the probability of the event
RELATIONSHIPS AMONG EVENTS
(not E): The event “E does not occur”
(A&B): The event “both A and B occur”
(AorB): The event “either A or B or both occur”
Chapter 4: Probability Concepts Page -8- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
RULES OF PROBABILITY
1. Addition Rule
if A & B are disjoint then
P(A B) = P(A) + P(B)
if A & B are not disjoint then
P(A B) = P(A) + P(B) - P(A&B)
2. Complementation Rule
( ) 1 ( )P A P A
3. P(S) = 1
4. P(Null) = 0
5. P(A or (not A)) = 1
6. A & B are independent iff (if and only if)
P(A&B) = P(A)*P(B) (Multiplication)
Chapter 4: Probability Concepts Page -9- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Example: Seatbelt Experiment
Given: P(Y) = 0.2 & P(N) = 0.8
Then the sample space of the responses of three people is:
S = {YYY, YYN, YNY, NYY, YNN, NYN, NNY, NNN}
The probability of each outcome using rule 6 is:
{0.008, 0.032, 0.032, 0.032, 0.128, 0.128, 0.128, 0.512}
Some sample probabilities of events
A = At least two wear seatbelts
P(A) = 0.008 + (3*0.032) = 0.104
B = Exactly two wear seatbelts
P(B) = (3*0.032) = 0.096
C = Exactly one wears a seatbelt
P(C) = (3*0.128) = 0.384
P(AUC) = P(A) + P(C)
= 0.104 + 0.384 = 0.488
P(A&C) = Empty Set = 0
P(A&B) = (3*0.032) = 0.096
P(AUB) = P(A) + P(B) - P(A&B)
= 0.104 + 0.096 - 0.096 = 0.104
not A = 1 - P(A)
= 1 - 0.014 = 0.896
Chapter 4: Probability Concepts Page -10- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Example Using Grouped Data Table
Classes Frequency CM Rel.Freq
910-929 1 919.5 0.033
930-949 1 939.5 0.033
950-969 3 959.5 0.100
970-989 9 979.5 0.300
990-1009 7 999.5 0.233
1010-1029 6 1019.5 0.200
1030-1049 2 1039.5 0.067
1050-1069 1 1059.5 0.033
TOTALS 30 1.000
Grouped Data Table
P(E) = (f / N)
Probability of x < 950
= (1 + 1) / 30 = 0.067
Probability of 950 <= x <= 1009
= (3 + 9 + 7) / 30 = 0.633
Chapter 4: Probability Concepts Page -11- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
CONTINGENCY TABLES (N-WAY GROUPING)
When you wish to group more than one variable you would use contingency tables. Examples: Sex by Age, Height by Weight, Sales by Product Line, etc.
Sample Contingency Table Sex\Age Under 21 21-25 Over 25 Total
Male Freq uency Counts #of Males
Female #of Females
Total #<21 #of21-25 #>25 GrandTotal
Cross-classified data is generally presented in contingency tables.
The numbers presented in the body of the table provide the
frequencies for various contingencies. These numbers are
presented in individual CELLS.
Chapter 4: Probability Concepts Page -12- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Example
Age & gender data for freshmen in a calculus class is presented.
The contingency table is shown below (yellow highlight is not part
of the table):
R1 R2 R3
21< 21-25 >25 Total
A1 Males 8 12 2 22
A2 Females 12 13 3 28
Total 20 25 5 50
INDIVIDUAL EVENTS
A1 = Event the student selected is male
A2 = Event the student selected is female
R1 = Event the student selected is under 21
R2 = Event the student selected is between 21-25
R3 = Event the student selected is over 25
NOTE:
A1 & A2 are mutually exclusive
R1, R2, & R3 are mutually exclusive
Chapter 4: Probability Concepts Page -13- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
JOINT EVENTS
(A1&R1): The student selected is a male under 21.
(A2&R3): The student selected is a female over 25.
In the above table there are: 2 x 3 = 6 joint events. That is, all
CELLS in the table are considered joint events and therefore joint
probabilities for each cell can be calculated.
Each joint event is also mutually exclusive from the other joint
event.
Probability of a joint event is calculated as:
P(A&B) = (f / N)
Where: f is the cell frequency.
N is the population size.
Example:
P(A1&R1) = 8/50 = 0.16 = 16%
P(A2&R3) = 3/50 = 0.06 = 6%
Chapter 4: Probability Concepts Page -14- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
MARGINAL PROBABILITY
From the row and column totals of a contingency table it is
possible to calculate marginal probabilities.
Example:
P(A1) = (Total of A1) / N
= 22 / 50
= 0.44
Note: The sum of all joint probabilities on a row or a column
equals the marginal probability in that row or column.
Chapter 4: Probability Concepts Page -15- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
CONDITIONAL PROBABILITY
The probability of an event given that another event has occurred.
Notation: P(A|B) = f/N
-or-
Generally: P(A|B) = P(A&B) / P(B)
Where: A is the event of interest.
B is the event that has occurred.
f is the conditional frequency.
N is the conditional population size.
Example:
Given that a student selected is male, what is the probability of
him being less than 21 years old?
P(R1|A1) = 8/22 = 0.36
-or-
P(R1|A1) = P(A1&R1) / P(A1)
= 0.16 / 0.44
= 0.36
Chapter 4: Probability Concepts Page -16- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
INDEPENDENCE
Recall: For any non-trivial (that is, the probability is not zero)
events independence is:
P(A&B) = P(A)P(B)
Further, two events are independent iff the following two
conditions are also true.
1) P(B|A) = P(B)
2) In terms of marginal & joint probabilities:
P(A&B) = P(A)P(B|A)
Example:
Using the card drawing experiment. Note: Once a card is picked it
is replaced back in the deck.
Events P(E)
A. A card selected is a king of diamond 1/52 = 0.019
B. A card selected is a king 4/52 = 0.077
C. A card selected is a diamond 13/52 = 0.25
D. A card selected is not a face card 40/52 = 0.769
Q: What is the probability of selecting a card that is the king of
diamond?
A: Since there is only one king of diamond in a deck of cards the
joint probability is calculated as:
P(B&C) = 1 / 52 = 0.019
Chapter 4: Probability Concepts Page -17- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Q: Are events B & C independent?
Standard formula if independent is assumed:
1) P(B&C) = P(B)P(C)
P(BC) = (0.077)(0.25) – calculated using (f/ N)
= 0.019
Additional Condition using marginal and joint probabilities
2a) Recall: P(C|B) = f / N = 1/4 = 0.25 = P(C)
That is, probability of getting a diamond, given that the card
selected is a king.
2b) Therefore,
P(B&C) = (0.077)(0.25) = 0.019
Where: 0.077 is computed using the (f/N); and 0.25 is calculated
as shown in (2a)
Chapter 4: Probability Concepts Page -18- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
MUTUALLY EXCLUSIVE V/S INDEPENDENT
When the occurrence of one event has NO effect on the probability of
occurrence of another event the two events are independent. That is,
P(A&B) = P(A)P(B)
When two events cannot occur simultaneously, they are mutually exclusive.
That is, P(A&B) = 0.
NOTE: If two events are independent they cannot be mutually exclusive,
and vice versa.
Example: Independent Events
Consider a fair coin and a fair six-sided die. Consider two events: A: Obtaining heads B: Rolling a 6. Then we can reasonably assume that events A and B are independent, because the outcome of one does not affect the outcome of the other. The probability that both A and B occur is: P(A&B) = P(A)P(B) = (1/2)(1/6) = 1/12. Since this value is not zero, then events A and B cannot be mutually exclusive.
Example: Mutually Exclusive Events
Consider a fair six-sided die as before, only in addition to the numbers 1 through 6 on each face, we have the property that the even-numbered faces are colored red, and the odd-numbered faces are colored green. Consider the two events: A: Rolling a green face, and B: Rolling a 6. Then, P(A) = 1/2 (two colors – red and green); and P(B) = 1/6 A and B cannot simultaneously occur, since rolling a 6 means the face is red, and rolling a green face means the number showing is odd. Therefore: P(A&B) = 0.
Chapter 4: Probability Concepts Page -19- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
BAYES' RULE (OPTIONAL) -- SKIP IN STA308
Exhaustive Events
Events A1, A2, ..., Ak are said to be exhaustive if at least one of the
events must occur when the experiment is performed.
Example 1:
Picking a student in the calculus class example implies that either
A1 or A2 is true. (A1 & A2 are exhaustive and mutually exclusive
events. That is, when the experiment is performed exactly one of
the events must occur).
Example 2:
In another experiment of selecting a card from a deck of 52 cards
the following events are defined:
A. A card selected is not a face card.
B. A card selected is a face card.
C. A card selected is a diamond
Events A, B, & C are exhaustive.
Events A & B are exhaustive and mutually exclusive.
Events (AUB) & C are not mutually exclusive because
event C is a subset of (AUB).
Chapter 4: Probability Concepts Page -20- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
The Rule Of Total Probability
If events A1, A2, A3, ..., Ak are mutually exclusive and exhaustive,
then for any event B,
P(B) = P(A1)P(B|A1) + P(A2)P(B|A2) + ... +
P(Ak)P(B|Ak).
Bayes' Rule
If events A1, A2, A3, ..., Ak are mutually exclusive and exhaustive,
then for any event B,
P(Ai|B) = P(Ai)P(B|Ai)
P(A1)P(B|A1) + P(A2)P(B|A2) + ...
+ P(Ak)P(B|Ak)
Where: Ai is any one of the events A1, A2, ..., Ak
Other Definitions:
Prior Probability: The probability of an event.
Posterior Probability: Conditional probability.
Chapter 4: Probability Concepts Page -21- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Example
The National Center for Health Statistics provides information on
suicides by sex and method used. The information published said
there were 30,904 total suicides in the US. of which 24,226 were
males and the remaining were females. The following table gives
the conditional probability of a male using a particular method
and a female using a particular method.
Methods Male (M) Female(F)
Poisoning (P) 0.145 0.377
Hanging (H) 0.155 0.127
Firearms (G) 0.641 0.395
Other (O) 0.059 0.101
Solution Given: Total suicides = 30,904 Male suicides = 24,226 Female suicides = 6,678 Therefore: P(M) = 24226/30904 = 0.784 P(F) = 6678/30904 = 0.216
Chapter 4: Probability Concepts Page -22- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Q: Determine the probability that a firearm was used for the
suicide. That is, what is P(G) ?
A: P(G) = P(M&G) + P(F&G)
= P(M)P(G|M) + P(F)P(G|F)
= (0.784)(0.641) + (0.216)(0.395)
= 0.58786
Q: What is the prior probability that the person is female?
A: P(F) = 0.216
Q: What is the posterior probability that the person was a female
given that a firearm was used?
A: P(F|G) = P(F&G)
P(G)
= P(F)P(G|F)
P(G)
= (0.216 * 0.395) / 0.587
= 0.145
Chapter 4: Probability Concepts Page -23- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
COUNTING RULES
The Fundamental Counting Rule
Suppose there are r actions to be performed in a definite order.
If there are m1 possibilities for the first action, m2 for the second
action, etc. Then the total number of possibilities is:
m1 m2 mr
Factorials
Let k be a positive integer. Then the product of the first k positive
integers is called k factorial and is denoted as:
k! = k(k - 1) 2 1
Examples:
3! = 3 2 1 = 6
0! = 1
1! = 1
Chapter 4: Probability Concepts Page -24- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Permutations
The number of possible permutations of r objects from a
collection of m objects is given by:
(m)r = m!
(m - r)!
Note: (m)m = m!
Example:
A zip code consists of five digits.
a) How many possible zip codes are there? b) How many possible zip codes are there in which no digit
appears more than once.
Solution Each location can get any of the 10 numbers selected one at a
time. That is, (m)r = (10)1 = 10.
There are five such locations, thus there are:
10 10 10 10 10 = 100,000 possible zip codes.
If no digit appears more than once then there are:
(m)r = (10)5
= 10 9 8 7 6
= 30,240 possible zip codes
Chapter 4: Probability Concepts Page -25- Class Notes to accompany: Introductory Statistics, 9
th Ed, By Neil A. Weiss
Prepared by: Nina Kajiji
Combinations
The number of possible combinations of r objects from a
collection of m objects.
m
r = m!
r!(m - r)!
Example:
How many samples of size 5 are possible from a population of size
70?
Solution
The # of samples of size 5 from a population of 70 is:
70
5 = 70!
5!65! = 12,103,014