probability
TRANSCRIPT
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Chap 2-1
Concepts & Applications
Basic Probability
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chap 2-2
Important Terms
Random Experiment – a process leading to an uncertain outcome
A coin is thrown
A consumer is asked which of two products he
or she prefers
The daily change in an index of stock market
prices is observed
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chap 2-3
3
Sample Space - Collection of all possible outcomes
e.g.: All six faces of a die:
e.g.: All 52 cardsa deck of bridge cards
Important Terms
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chap 2-4
Event – any subset of basic outcomes from the sample space
Important Terms
Simple Event
Outcome With 1 Characteristic
Red card from a deck of bridge cards
Ace card from a deck of bridge cards
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Joint Event
2 Events Occurring Simultaneously
A and B, (AB):
Red, ace card from a bridge deck
Male, over age 20
Important Terms
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chap 2-6
Compound Event
One or Another Event Occurring
D or E, (DE): Ace or Red card from bridge deck
Important Terms
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chap 2-7
Important Terms
Intersection of Events – If A and B are two events in a sample space S, then the intersection, A ∩ B, is the set of all outcomes in S that belong to both A and B
(continued)
A BAB
S
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chap 2-8
Important Terms
A and B are Mutually Exclusive Events if they have no basic outcomes in common i.e., the set A ∩ B is empty
(continued)
A B
S
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chap 2-9
Important Terms
Union of Events – If A and B are two events in a sample space S, then the union, A U B, is the set of all outcomes in S that belong to either
A or B
(continued)
A B
The entire shaded area represents A U B
S
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chap 2-10
Important Terms
Events E1, E2, … Ek are Collectively Exhaustive events if E1 U E2 U . . . U Ek = S i.e., the events completely cover the sample space
The Complement of an event A is the set of all basic outcomes in the sample space that do not belong to A. The complement is denoted
(continued)
A
AS
A
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chap 2-11
Examples
Let the Sample Space be the collection of all possible outcomes of rolling one die:
S = [1, 2, 3, 4, 5, 6]
Let A be the event “Number rolled is even”
Let B be the event “Number rolled is at least 4”
Then
A = [2, 4, 6] and B = [4, 5, 6]
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chap 2-12
(continued)
Examples
S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6]
5] 3, [1, A
6] [4, BA
6] 5, 4, [2, BA
S 6] 5, 4, 3, 2, [1, AA
Complements:
Intersections:
Unions:
[5] BA
3] 2, [1, B
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chap 2-13
Mutually exclusive: A and B are not mutually exclusive
The outcomes 4 and 6 are common to both
Collectively exhaustive: A and B are not collectively exhaustive
A U B does not contain 1 or 3
(continued)
Examples
S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6]
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chap 2-14
Probability
Probability – the chance that an uncertain event will occur (always between 0 and 1)
0 ≤ P(A) ≤ 1 For any event A
Certain
Impossible
.5
1
0
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chap 2-15
Assessing Probability
There are three approaches to assess the probability of an uncertain event:
1. classical probability
Assumes all outcomes in the sample space are equally likely to
occur
spacesampletheinoutcomesofnumbertotal
eventthesatisfythatoutcomesofnumber
N
NAeventofyprobabilit A
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chap 2-16
Assessing Probability
Three approaches (continued)
2. relative frequency probability
the limit of the proportion of times that an event A occurs in a large number of trials, n
3. subjective probability
an individual opinion or belief about the probability of occurrence
populationtheineventsofnumbertotal
Aeventsatisfythatpopulationtheineventsofnumber
n
nAeventofyprobabilit A
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chap 2-17
Probability Postulates
1. If A is any event in the sample space S, then
2. Let A be an event in S, and let Oi denote the basic outcomes. Then
(the notation means that the summation is over all the basic outcomes in A)
3. P(S) = 1
1P(A)0
)P(OP(A)A
i
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chap 2-18
Probability Rules
The Complement rule:
The Addition rule: The probability of the union of two events is
1)AP(P(A)i.e., P(A)1)AP(
B)P(AP(B)P(A)B)P(A
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chap 2-19
Addition Rule Example
Consider a standard deck of 52 cards, with four suits:
♥ ♣ ♦ ♠
Let event A = card is an Ace
Let event B = card is from a red suit
Find P(A or B)?
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chap 2-20
P(Red U Ace) = P(Red) + P(Ace) - P(Red ∩ Ace)
= 26/52 + 4/52 - 2/52 = 28/52Don’t count the two red aces twice!
BlackColor
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
(continued)
Addition Rule Example
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chap 2-21
Counting the Possible Outcomes
Use the Combinations formula to determine the number of combinations of n things taken k at a time
where n! = n(n-1)(n-2)…(1) 0! = 1 by definition
k)!(nk!
n! Cn
k
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chap 2-22
Example
The sample space contains 5 A’s and 7 B’s. What is the probability that a randomly selected set of 2 will include 1 A and 1 B?
Sol:
53.066
35
66
75)1&1(
)1&1(122
71
51
BAP
C
CCBAP
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chap 2-23
Practice Questions
Q # 1: The sample space contains 6 red and 4 green marbles. What is the probability that a randomly selected set of 3 will include 1 Red & 2 Green marbles.
Ans: 36/120Q # 2: ABC Inc. is hiring managers to till four key positions. The candidates are five men and three women. Assuming that every combination of men and women is equally likely to be chosen, what is the probability that at least one woman will be selected?
Ans: 13/14Hint: A: at least one woman is selected
Use complement law
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chap 2-24
A conditional probability is the probability of one event, given that another event has occurred:
P(B)
B)P(AB)|P(A
P(A)
B)P(AA)|P(B
The conditional probability of A given that B has occurred
The conditional probability of B given that A has occurred
Conditional Probability
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chap 2-25
What is the probability that a car has a CD player, given that it has AC ?
i.e., we want to find P(CD | AC)
Conditional Probability Example
Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.
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chap 2-26
Conditional Probability Example
No CDCD Total
AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0
Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.
.2857.7
.2
P(AC)
AC)P(CDAC)|P(CD
(continued)
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chap 2-27
Conditional Probability Example
No CDCD Total
AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0
Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is 28.57%.
.2857.7
.2
P(AC)
AC)P(CDAC)|P(CD
(continued)
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chap 2-28
Multiplication Rule
Multiplication rule for two events A and B:
also
P(B)B)|P(AB)P(A
P(A)A)|P(BB)P(A
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Question
chap 2-29
If a card is selected, at random, from a deck of 52
cards, what is the probability that the card is an ace
of red color?
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chap 2-30
Multiplication Rule Example
P(Red ∩ Ace) = P(Red| Ace)P(Ace)
BlackColor
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
52
2
52
4
4
2
52
2
cards of number total
ace and red are that cards of number
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chap 2-31
Statistical Independence
Two events are statistically independent if and only if:
Events A and B are independent when the probability of one event is not affected by the other event
If A and B are independent, then
P(A)B)|P(A
P(B)P(A)B)P(A
P(B)A)|P(B
if P(B)>0
if P(A)>0
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chap 2-32
Statistical Independence Example
No CDCD Total
AC .2 .5 .7
No AC .2 .1 .3
Total .4 .6 1.0
Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.
Are the events AC and CD statistically independent?
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• Out of a target audience of 2,000,000, ad Out of a target audience of 2,000,000, ad A A reaches reaches 500,000 viewers, 500,000 viewers, BB reaches 300,000 viewers and both reaches 300,000 viewers and both ads reach 100,000 viewers.ads reach 100,000 viewers.
• What is What is PP((AA | | BB)?)?
500,000( ) .25
2,000,000P A
300,000( ) .15
2,000,000P B
100,000( ) .05
2,000,000P A B
Statistical IndependenceStatistical IndependenceStatistical IndependenceStatistical Independence
Example: Television AdsExample: Television Ads
( ) .05( | ) .30
( ) .15
P A BP A B
P B
.3333 or 33%.3333 or 33%
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• So, So, PP(ad (ad AA) = .25) = .25 PP(ad (ad BB) = .15) = .15 PP((AA BB) = .05) = .05 PP((AA | | BB) = .3333 ) = .3333
• PP((AA | | BB) = .3333 ≠ ) = .3333 ≠ PP((AA) = .25) = .25
• PP((AA))PP((BB)=(.25)(.15)=.0375 ≠ )=(.25)(.15)=.0375 ≠ PP((AA BB)=.05 )=.05
• Are events Are events AA and and BB independent? independent?
Independent EventsIndependent EventsIndependent EventsIndependent Events
Example: Television AdsExample: Television Ads
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chap 2-35
Q:1 A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? of rolling an odd number?Q:2 A single 6-sided die is rolled. What is the probability of rolling a 2 or a 5?Q:3 A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble? a green marble? a blue marble? a yellow marble?Q:4 A glass jar contains 1 red, 3 green, 2 blue, and 4 yellow marbles. If a single marble is chosen at random from the jar, what is the probability that it is yellow or green?
Practice Questions
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chap 2-36
Q:5 On New Year's Eve, the probability of a person having a car accident is 0.09. The probability of a person driving while intoxicated is 0.32 and probability of a person having a car accident while intoxicated is 0.15. What is the probability of a person driving while intoxicated or having a car accident?Q:6 A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a card that is not a king?Q:7 A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and an eight?
Practice Questions
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Q # 8: A hamburger chain found that 75% of all customers use mustard, 80% use ketchup, and 65% use both.
a.What is the probability that a customer will use at least one of these? (0.90)
b. What is the probability that a ketchup user uses mustard? (0.8125)
Q # 9: The probability of A is 0.70 and the probability of B is 0.80 and the probability of both is 0.50. What is the conditional probability of A given B? Are A and B statistically independent?
Practice Questions
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chap 2-38
Practice Questions
Q # 10: (a) A fair die is rolled twice. Let A be the event of an odd total and B the event of an ace on the first die. Verify that P(A/B) = P(A) (b) The probability that a man will be alive in 25 years is 3/5 and that his wife will be alive in 25 years is 2/3. Find the probability that (i) both will be alive (ii) At least one will be alive.
Q # 11: A music store owner finds that 25% of the customers entering the store ask as assistant for help and 30% of the customers make a purchase before leaving. It is also found that 20% of all customers both ask for assistance and make a purchase. What is the probability that a customer does at least one of these two things?
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Q # 12: The accompanying table shows proportions of adults categorized to whether they are readers or non-readers of newspapers and whether or not they voted in the last election.
Practice Questions
Voted Readers Non-readers
Yes 0.63 0.13
No 0.14 0.10
What is the probability that a randomly chosen adult from this population who did not read newspaper did not vote? (0.4348)
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chap 2-40
Q # 13: In a certain college, 4% of the men and 1% of the women are taller than 6 feet. Furthermore 60% of the students are women. Now, if a student is selected at random and is taller than 6 feet, what is the probability that the student is a woman?
Practice Questions