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DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS

Introduction to Business Statistics Introduction to Business Statistics QM 120 Ch t 4 Chapter 4

Spring 2008

• Dr. Mohammad Zainal

Chapter 4: Experiment, outcomes, and sample space

2

Probability and statistics are related in an important way. It

is used to allow us to evaluate the reliability of our

2

conclusions about the population when we have only sample information.

Data are obtained by observing either uncontrolled events

in nature or controlled situation in laboratory. We use the term experiment to describe either method

• f

data term experiment to describe either method

• f

data collection.

The

• bservation
• r

measurement generated by an e

• bse vatio
• easu e

e t ge e ated by a experiment may or may not produce a numerical value. Here are some examples:

Recording a test grade Interviewing a householder to obtain his or her opinion in certain

issue issue.

QMIS 120, CH 4

Chapter 4: Experiment, outcomes, and sample space

3

Table 1: Examples of experiments, outcomes, and sample spaces

Sample Space Outcomes Experiment

3

S = {Head, Tail} Head, Tail Toss a coin once S = {1, 2, 3, 4, 5, 6} 1, 2, 3, 4, 5, 6 Roll a die once S = {Win, Lose} Win, Lose Play a lottery S = {M,F} M, F Select a student S = {HH HT TH TT} HH HT TH TT Toss a coin twice

Venn diagram is a picture that depicts all possible outcomes

S {HH, HT, TH, TT} HH, HT, TH, TT Toss a coin twice

for an experiment while in tree diagram, each outcome is represented a branch of a tree.

• H
• T

Venn Diagram

H T

Tree Diagram

T

QMIS 120, CH 4

Chapter 4: Experiment, outcomes, and sample space

4

Example: Draw the Venn and tree diagrams for the experiment of tossing a coin twice.

4

Solution:

E a le D a a t ee dia a fo th ee to e of a oi Li t Example: Draw a tree diagram for three tosses of a coin. List all outcomes for this experiment in a sample space S.

Solution

QMIS 120, CH 4

Chapter 4: Experiment, outcomes, and sample space

5

A simple event is the outcome that is observed on a single

repetition of the experiment. It is often denoted by E with a

5

p p y subscript. Example: Toss a die and observe the number that appears on Example: Toss a die and observe the number that appears on the upper face. List the simple event in the experiment.

Solution: Solution:

QMIS 120, CH 4

Chapter 4: Experiment, outcomes, and sample space

6

We can now define an event (or compound event) as a

collection of simple events, often denoted a capital letter.

6

Example: Tossing a die (continued) We can define the events A and B as follow, A: Observe an odd number B: Observe a number less than 4 S l ti Solution:

QMIS 120, CH 4

Chapter 4: Experiment, outcomes, and sample space

7

Example: Suppose we randomly select two persons from members of a club and observe whether the person selected is

7

a man or woman. Write all the outcomes from experiment. Draw the Venn and tree diagrams for this experiment.

Solution: Solution:

Two events are mutually exclusive if, when one event

• ccurs, the other cannot, and vice versa.

The set of all simple events is called the sample space, S.

QMIS 120, CH 4

Chapter 4: Calculating Probability

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Probability is a numerical measure of the likelihood that an

event will occur.

8

• Two properties of probability

The probability of an even always lies in the range 0 to 1 The probability of an even always lies in the range 0 to 1. 0 ≤ P(Ei) ≤ 1 0 ≤ P(A) ≤ 1 0 ≤ P(A) ≤ 1

The sum of the probabilities of all simple events for an experiment,

denoted by ΣP(Ei), is always 1. y ) y

∑ P(Ei) = P(E1) + P(E2) + P(E3) + . . . . = 1

QMIS 120, CH 4

Chapter 4: Calculating Probability

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Three conceptual approaches to probability 1 Classical probability

9

• 1. Classical probability

Two or more events that have the same probability of occurrence are

said to be equally likely events.

The probability of a simple event is equal to 1 divided by the total

number of all final outcomes for an equally likely experiment.

Classical probability rule to find probability

1 ( ) P E = ( ) Total number of outcomes Number of outcomes favorable to ( ) Total number of outcomes

i

P E A P A = =

QMIS 120, CH 4

Chapter 4: Calculating Probability

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Example: Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin

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probability of obtaining a tail for one toss of a coin.

Solution

QMIS 120, CH 4

Chapter 4: Calculating Probability

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Example: Find the probability of obtaining an even number in

• ne roll of a die.

11

Solution

QMIS 120, CH 4

Chapter 4: Calculating Probability

12

Calculating the probability of an event:

List all the simple events in the sample space.

12

List all the simple events in the sample space. Assign an appropriate probability to each simple event. Determine which simple events result in the event of interest. Determine which simple events result in the event of interest. Sum the probabilities of the simple events that result in the event of

interest. Always

• 1. Include all simple events in

the sample space.

• 2. Assign realistic probabilities

to the simple events.

QMIS 120, CH 4

Chapter 4: Calculating Probability

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Example: A six years boy has a safe box that contains four banknotes: One‐Dinar, Five‐Dinar, Ten‐Dinar, Twenty‐Dinar.

13

His sister which is a three years old girl randomly grabbed three banknotes from the safe box to buy a 30 KD toy. Find the

• dds (probability) that this girl can buy the toy
• dds (probability) that this girl can buy the toy.

Solution

QMIS 120, CH 4

Chapter 4: Calculating Probability

14

• 2. Relative frequency concept of probability

Suppose we want to know the following probabilities:

14

Suppose we want to know the following probabilities:

The next car coming out of an auto factory is a “lemon” A randomly selected family owns a home A randomly selected woman has never smoked

The outcomes above are neither equally likely nor fixed for each

l sample.

The variation goes to zero as n becomes larger and larger

If i i d i d A i b d f

If an experiment is repeated n times and an event A is observed f

times, then

( ) f P A n =

QMIS 120, CH 4

Chapter 4: Calculating Probability

15

Example: In a group of 500 women, 80 have played golf at least once. Suppose one of these 500 woman is selected. What

15

is the probability that she played golf at least once

Solution

QMIS 120, CH 4

Calculating Probability

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Example: Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons. What is the

16

y probability that the next car manufactured at that factory is a lemon?

S l i Solution:

QMIS 120, CH 4

Calculating Probability

17

Example: Lucca Tool Rental would like to assign probabilities to the number of car polishers it rents each day. Office records

17

p y show the following frequencies of daily rentals for the last 40 days.

Number of Polishers Rented Number of Days 4 1 6 2 18 2 18 3 10 4 2

QMIS 120, CH 4

Calculating Probability

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Law of large numbers: If an experiment is repeated again

and again, the probability of an event obtained from the

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relative frequency approaches the actual probability.

• 3. Subjective probability

Suppose we want to know the following probabilities:

pp g p

A student who is taking a statistics class will get an A grade. KSE price index will be higher at the end of the day. China will dominate the gold medal list in the 2008 Olympics.

Subjective probability is the probability assigned to an event based on

subjective judgment experience information and belief subjective judgment, experience, information, and belief.

QMIS 120, CH 4

Counting Rule

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Suppose that an experiment involves a large number N of

simple events and you know that all the simple events are

19

equally likely. Then each simple event has probability 1/N and the probability of an event A can be calculated as

( ) Where is the number of simple events that result in event A

A A

n P A N n =

The mn rule

Consider an experiment that is performed in two stages. If the first

stage can be performed in m ways and for each of these ways, the second stage can be accomplished in n ways, then there mn ways to accomplish the experiment

QMIS 120, CH 4

Counting Rule

20

Example: Suppose you want to

• rder a car in one of three

20

styles and in one of four paint

• colors. To find out how many
• ptions are available you can
• ptions are available, you can

think of first picking one of the m = 3 styles and then one of n = 4 colors.

QMIS 120, CH 4

Counting Rule

21

The extended mn rule

If an experiment is performed in k stages, with n1 ways to accomplish

21

If an experiment is performed in k stages, with n1 ways to accomplish

the first stage, n2 to accomplish the second stage…, and nk ways to accomplish the kth stage, the number of ways to accomplish the experiment is p

1 2 3... k

n n n n

Example: A bus driver can take three routes from city A to city B, four routes from city B to city C, and three routes from city C t it D F t li f A t D th d i t d i C to city D. For traveling from A to D, the driver must drive from A to B to C to D, how many possible routes from A to D are available are available

QMIS 120, CH 4

Counting Rule

22

Example: Ahmad has invested in two stocks, ARC Oil and Coal Mining. Ahmad has determined that the possible

22

g p

• utcomes of these investments three months from now are as

follows.

Investment Gain or Loss in 3 Months (in €000)

Arc Oil Coal Mining 10 5 8

• 2
• 20

QMIS 120, CH 4

Counting Rule

23

Solution

23

QMIS 120, CH 4

Counting Rule

24 24

QMIS 120, CH 4

Counting Rule

25

A counting rule for permutations (orderings)

The number of ways we can arrange n distinct objects, taking them

25

The number of ways we can arrange n distinct objects, taking them

r at a time, is

!

n

n P P = = ( )! where ! ( 1)( 2).....(3)(2)(1) and 0! 1

r n r

P P n r n n n n = = − = − − =

A counting rule for combination

The number of distinct combinations of n distinct objects that can The number of distinct combinations of n distinct objects that can

be formed, taking them r at a time, is

! n! ( ) !( )!

n n n r r r

n C C r n r = = = −

QMIS 120, CH 4

Counting Rule

26

Permutations: Given that position (order) is important, if

• ne has 5 different objects (e.g. A, B, C, D, and E), how

26

many unique ways can they be placed in 3 positions (e.g. ADE, AED, DEA, DAE, EAD, EDA, ABC, ACB, BCA, BAC etc ) etc.)

Combinations: If one has 5 different objects (e.g. A, B, C, D,

d E) h th b d 3 bj t and E), how many ways can they be grouped as 3 objects when position does not matter (e.g. ABC, ABD, ABE, ACD, ACE, ADE are correct but CBA is not ok as is equal to ABC) , q )

QMIS 120, CH 4

Marginal & Conditional Probabilities

27

Suppose all 100 employees of a company were asked

whether they are in favor of or against paying high salaries

27

y g p y g g to CEOs of U.S. companies. The following table gives a two‐ way classification of their responses.

Total Against In favor 60 45 15 Male 40 36 4 Female Total 100 81 19

Marginal probability is the probability of a single event

without consideration of any other event y

QMIS 120, CH 4

Marginal & Conditional Probabilities

28

Suppose one employee is selected, he/she maybe classified

either on the bases of gender alone or on the bases of

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• pinion.

The probability of each of the following event is called

marginal probability

P(male) =

Total Against In favor

P(female) = P(in favor) =

g 60 45 15 40 36 4 Male Female T t l

P(against) =

Suppose the employee selected is known to be male. What

i h b bili h h i i f ?

Total 100 81 19

is the probability that he is in favor?

QMIS 120, CH 4

Marginal & Conditional Probabilities

29

This probability is called conditional probability and is

written as “the probability that the employee selected is in

29

favor given that he is a male.” P( in favor | male )

This event has already occurred Read as “given” The event whose probability is to be determined

Conditional probability is the probability that an even will

• ccur given that another event has already occurred. If A

and B are two events then the conditional probability of A and B are two events, then the conditional probability of A given B is written as P(A | B ) P(A | B )

QMIS 120, CH 4

Marginal & Conditional Probabilities

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Example: Find the conditional probability P(in favor | male) for the data on 100 employees

30

p y

Solution

QMIS 120, CH 4

Marginal & Conditional Probabilities

31

Example: Find the conditional probability P(female|in favor) for the data on 100 employees

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p y

Solution

QMIS 120, CH 4

Marginal & Conditional Probabilities

32

Example: Consider the experiment of tossing a fair die. Denote by A and B the following events:

32

y g A={Observing an even number}, B={Observing a number of dots less than or equal to 3}. Find the probability of the event A i e the e e t B A, given the event B.

Solution

QMIS 120, CH 4

Mutually Exclusive Events

33

Events that cannot occur together are called mutually

exclusive events.

33

Example: Consider the following events for one roll of a die A b i b d {2 4 6} 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} C: a number less than 5 is observed {1,2,3,4} ‐ A and B are mutually exclusive events but A and C are not. ‐ How about B and C? ‐ Simple events are mutually exclusive always.

QMIS 120, CH 4

Independent vs. Dependent events

34

Two events are said to be independent if the occurrence of

• ne does not affect the probability of the other one.

34

A and B are said to be independent events if either

P(A| B) = P(A) or P(B | A) = P(B) P(A| B) P(A) or P(B | A) P(B) Example: Refer to the information on 100 employees. Are p p y events “female (F)” and “in favor (A)” independent?

Solution:

QMIS 120, CH 4

Independent vs. Dependent events

35

What is the difference between mutually exclusive and

independent events?

35

It is common to get confused or not to tell the difference between

these two terminologies.

When two events are mutually exclusive, they cannot both happen.

Once the event B has occurred, event A cannot occur, so that P(A|B) = 0, or vice versa. The occurrence of event B certainly affects the probability that event A can occur. Therefore,

Mutually exclusive events must be dependent. Independent events are never mutually exclusive Independent events are never mutually exclusive.

But, dependent events may or may not be mutually exclusive

QMIS 120, CH 4

Independent vs. Dependent events

36

Example: A sample of 420 people were asked if they smoke or not and whether they are graduate or not. The following two‐way

36

classification table gives their responses

Not a college graduate College graduate Not a college graduate College graduate 80 35 Smoker 175 130 Nonsmoker

If an person is selected at random from this sample, find the

probability that this person is a

College graduate (G). Nonsmoker (NS). Smoker (S) given the person is not a college graduate (NG). College graduate (G) given the person is a nonsmoker (NS). Are the events Smoker and college graduate independent? Are the events Smoker and college graduate independent?

Why?

QMIS 120, CH 4

Independent vs. Dependent events

37 37

QMIS 120, CH 4

Complementary Events

38

Two mutually exclusive events that taken together include

all the outcome for an experiment (sample space, S) are

38

called complementary events.

Consider the following Venn diagram:

The complement of event A, denoted by A and read as “A bar”

• r

“A

A A

A

A and read as A bar

• r

A complement,” is the event that includes all the outcomes for an experiment that

A

B

A

B

Since two complementary events, taken together, include all

the sample space S the sum of probabilities of all outcomes are not in A

Venn diagram of two complementary evens

the sample space S, the sum of probabilities of all outcomes is 1 P(A) + P(A) = 1 → P(A) = 1 ‐ P(A) P(A) + P(A) 1 → P(A) 1 P(A)

QMIS 120, CH 4

Complementary Events

39

Example: In a lot of five machines, two are defective. If one of machines is randomly selected, what are the complementary

39

events for this experiment and what are their probabilities?

Solution

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

40

Intersection of events

• Let A and B be two events defined in a sample space The

40

• Let A and B be two events defined in a sample space. The

intersection of A and B represents the collection of all

• utcomes that are common to both A and B is denoted by

f h f ll i any of the followings A and B, A ∩ B, or AB Example: Let A = the event that a Let A = the event that a person owns a PC Let B = the event that a

A B A and B

Intersection of events A and B

person owns a mobile

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

41

Multiplication rule

Sometimes we may need to find the probability of two

41

Sometimes we may need to find the probability of two

events happening together.

The probability of the intersection of two events A and B is The probability of the intersection of two events A and B is

called their joint probability. It is written P(AB) or P(A ∩ B) ( ) ( )

It can be obtained by multiplying the marginal probability

• f one event with the conditional probability of the second

p y

• ne.

Multiplication rule: The probability of the intersection of

two events A and B is P(AB) = P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B)

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

42

Example: The following table gives the classification of all employees of a company by gender and college degree.

42

p y p y y g g g

Total Not a college graduate (N) College graduate (G) 27 20 7 Male (M) 13 9 4 Female (F) 40 29 11 Total

If one employee is selected at random, what is the probability that the employee is a female and a college graduate?

40 29 11 Total

that the employee is a female and a college graduate?

Solution

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

43 43

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

44

Example: A box contains 20 DVD, 4 of which are defective. If two DVDs are selected at random (without replacement),

44

( p ) what is the probability that both are defective?

Solution:

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

45

If events A and B are independent, their joint probability

simplifies from

45

P(A ∩ B) = P(A)P(B |A) TO P(A ∩ B) = P(A)P(B )

Sometimes we know the joint probability of two events A

and B in this case the conditional probability of B given A and B, in this case, the conditional probability of B given A

• r A given B is

( ) ( | ) ( ) P A B P A B P B ∩ = ( ) ( | ) ( ) P A B P B A P A ∩ =

given that P( ) 0 and P( ) A B ≠ ≠

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

46

Example: According to a survey, 60% of all homeowners owe money on home mortgages. 36% owe money on both home

46

y g g y mortgages and car loans. Find the conditional probability that a homeowner selected at random owes money on a car loan i e that he o e

• ey o

a ho e

• t a e

given that he owes money on a home mortgage.

Solution:

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

47

Example: A computer company has two quality control inspectors, Mr. Smith and Mr. Robertson, who independently

47

p p y inspect each computer before it is shipped to a client. The probability that Mr. Smith fails to detect a defective PC is .02 hile it i 01 fo M Robe t o Fi d the

• bability that both

while it is .01 for Mr. Robertson. Find the probability that both inspectors will fail to detect a defective PC.

Solution: Solution:

QMIS 120, CH 4

Intersection of Events & Multiplication Rule

48

Example: The probability that a patient is allergic to Penicillin is .2. Suppose this drug is administrated to three patients.

48

pp g p Find a) The probability that all three of them are allergic to it b) At lea t o e of the i

• t alle

i b) At least one of them is not allergic

Solution:

QMIS 120, CH 4

49

Union of Events & Addition Rule

Union of events

• Let A and B be two events defined in a sample space S The

49

• Let A and B be two events defined in a sample space S. The

union of events A and B is the collection of all outcomes that belong either to A and B or to both A and B and is denoted b “ A B” “A B” by “ A or B” or “A ∪ B”

S A B

QMIS 120, CH 4

50

Union of Events & Addition Rule

Example: A company has 1000 employees. Of them, 400 are females and 740 are labor union members. Of the 400 females,

50

250 are union members. Describe the union of events “female” and “union member”

Solution

QMIS 120, CH 4

51

Union of Events & Addition Rule

Addition rule

• The probability of the union of two events A and B is

51

• The probability of the union of two events A and B is

P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Example: A university president has proposed that all students Example: 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 y y asked about their opinion on this issue and it shown in the following table.

Opinion Total Neutral Oppose Favor 70 10 15 45 Faculty

d h b b l h l d f l b

y 230 30 110 90 Student 300 40 125 135 Total

Find the probability that a person selected is a faculty member or in favor.

QMIS 120, CH 4

52

Union of Events & Addition Rule

Solution:

52

QMIS 120, CH 4

53

Union of Events & Addition Rule

Example: There are a total of 7225 thousand persons with multiple jobs in the US. Of them, 4115 thousand are male, 1742

53

p j thousand are single, and 905 thousand are male and single. What is the probability that a selected person is a male or i le? single?

Solution:

QMIS 120, CH 4

54

Union of Events & Addition Rule

The probability of the union of two mutually exclusive

events A and B is

54

P(A ∪ B) = P(A) + P(B) Example: A university president has proposed that all Example: 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 and it shown in the following table.

Opinion Total Neutral Oppose Favor 70 10 15 45 Faculty

Fi d th b bilit th t l t d i i f i

230 30 110 90 Student 300 40 125 135 Total

Find the probability that a person selected is in favor or is neutral

QMIS 120, CH 4

55

Union of Events & Addition Rule

55

Solution:

QMIS 120, CH 4

56

Union of Events & Addition Rule

Example: Eighteen percent of the working lawyers in the United States are female. Two lawyers are selected at random

56

y and it is observed whether they are male or female. a) Draw a tree diagram for this experiment b) Fi d the

• bability that at lea t o e of the t
• la

ye i a b) Find the probability that at least one of the two lawyers is a female.

Solution: Solution:

QMIS 120, CH 4

Law of Total Probability & Bayes’ Rule

57

Consider the following Venn diagram

B

57

B1 B2 Bn-1 Bn

B A

A1 A2 An-1 An

Can we find the area (probability) of A assuming that we know the probability of each Bi & P(A|B ) ? know the probability of each Bi & P(A|Bi) ?

QMIS 120, CH 4

58

Law of Total Probability & Bayes’ Rule

Addition Law

( ) ( ) ( ) ( ) P A B P A P B P A B ∪ + ∩

58

Multiplication Law

( ) ( ) ( ) ( ) P A B P A P B P A B ∪ = + − ∩

Multiplication Law

( ) ( ) ( | ) P A B P B P A B ∩ = ( ) ( ) ( | ) P A B P A P B A ∩ =

Conditional Probability

( ) ( | ) ( ) P A B P A B P B ∩ = ( ) ( | ) ( ) P A B P B A P A ∩ =

QMIS 120, CH 4

59

Law of Total Probability & Bayes’ Rule

( ) ( | ) P A P B A

Bayes’ theorem (two‐event case)

59

1 1 1 1 1 2 2

( ) ( | ) ( | ) ( ) ( | ) ( ) ( | ) P A P B A P A B P A P B A P A P B A = +

2 2 2

( ) ( | ) ( | ) ( ) ( | ) ( ) ( | ) P A P B A P A B P A P B A P A P B A = +

Bayes’ theorem

1 1 2 2

( ) ( | ) ( ) ( | ) P A P B A P A P B A +

Bayes theorem

1 1 2 2

( ) ( | ) ( | ) ( ) ( | ) ( ) ( | ) ... ( ) ( | )

i i i n n

P A P B A P A B P A P B A P A P B A P A P B A = + + +

QMIS 120, CH 4

60

Law of Total Probability & Bayes’ Rule

Example: Manufacturing firm that receives shipment of parts from two different suppliers. Currently, 65 percent of the

60

pp y p parts purchased by the company are from supplier 1 and the remaining 35 percent are from supplier 2. Historical Data u e t the uality ati

• f the t
• u

lie a e ho i the suggest the quality rating of the two supplier are shown in the table:

Good Parts Bad Parts Supplier 1 98 2 Supplier 1 95 5

a) Draw a tree diagram for this experiment with the probability of all outcomes probability of all outcomes b) Given the information the part is bad, What is the probability the part came from supplier 1?

QMIS 120, CH 4

61

Law of Total Probability & Bayes’ Rule

Solution:

61

QMIS 120, CH 4

62

Law of Total Probability & Bayes’ Rule

Example: An insurance company rents 35% of the cars for its customers from Avis and the rest from Hertz. From past

62

records they know that 8% of Avis cars break down and 5% of Hertz cars break down. A customer calls and complains that his rental car broke down What is the probability that his car his rental car broke down. What is the probability that his car was rented from Avis?

QMIS 120, CH 4

63

Law of Total Probability & Bayes’ Rule

Solution:

63

QMIS 120, CH 4