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6 Independence, Conditional Probability, and the Multiplication Rule

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150

CHAPTER 4 PROBABILITY AND PROBABILITY DISTRIBUTIONS

EXAMPLE

4.19

In a color preference experiment, eight toys are placed in a container. The toys

are identical except for color—two are red, and six are green. A child is asked to

choose two toys at random. What is the probability that the child chooses the two red

toys?

You can visualize the experiment using a tree diagram as shown in

Figure 4.13. Deﬁne the following events:

Solution

R: Red toy is chosen

G: Green toy is chosen

FI GU R E 4 .1 3

Tree diagram for

Example 4.19

First choice

Second choice

Red (1/7)

Simple event

RR

Red (2/8)

Green (6/7)

Red (2/7)

RG

GR

Green (6/8)

Green (5/7)

GG

The event A (both toys are red) can be constructed as the intersection of two events:

A ϭ (R on ﬁrst choice) ʝ (R on second choice)

Since there are only two red toys in the container, the probability of choosing red on

the ﬁrst choise is 2/8. However, once this red toy has been chosen, the probability of

red on the second choice is dependent on the outcome of the ﬁrst choice (see Figure 4.13). If the ﬁrst choice was red, the probability of choosing a second red toy is

only 1/7 because there is only one red toy among the seven remaining. If the ﬁrst

choice was green, the probability of choosing red on the second choice is 2/7 because

there are two red toys among the seven remaining. Using this information and the

Multiplication Rule, you can ﬁnd the probability of event A.

P(A) ϭ P(R on ﬁrst choice ʝ R on second choice)

ϭ P(R on ﬁrst choice) P(R on second choice)͉R on ﬁrst)

΂ ΃΂ ΃

2 1

2

1

ϭ ᎏᎏ ᎏᎏ ϭ ᎏᎏ ϭ ᎏᎏ

8 7

56

28

Sometimes you may need to use the Multiplication Rule in a slightly different form,

so that you can calculate the conditional probability, P(AͦB). Just rearrange the terms

in the Multiplication Rule.

4.6 INDEPENDENCE, CONDITIONAL PROBABILITY, AND THE MULTIPLICATION RULE

151

CONDITIONAL PROBABILITIES

The conditional probability of event A, given that event B has occurred is

P(A ʝ B)

P(A͉B) ϭ ᎏᎏ if P(B) 0

P(B)

The conditional probability of event B, given that event A has occurred is

P(A ʝ B)

P(B͉A) ϭ ᎏᎏ

P(A)

if

P(A)

0

Suppose that in the general population, there are

51% men and 49% women, and that the proportions of colorblind men and women

are shown in the probability table below:

Colorblindness, continued

Men(B)

Women (BC)

Total

Colorblind (A)

Not Colorblind (AC)

.04

.47

.002

.488

.042

.958

Total

.51

.49

1.00

If a person is drawn at random from this population and is found to be a man (event

B), what is the probability that the man is colorblind (event A)? If we know that the

event B has occurred, we must restrict our focus to only the 51% of the population

that is male. The probability of being colorblind, given that the person is male, is 4%

of the 51%, or

P(A ʝ B)

.04

P(A͉B) ϭ ᎏᎏ ϭ ᎏᎏ ϭ .078

P(B)

.51

What is the probability of being colorblind, given that the person is female? Now we

are restricted to only the 49% of the population that is female, and

P(A ʝ BC)

.002

P(A͉BC) ϭ ᎏᎏ

ϭ ᎏᎏ ϭ .004

P(BC)

.49

Notice that the probability of event A changed, depending on whether event B occured.

This indicates that these two events are dependent.

When two events are independent—that is, if the probability of event B is the same,

whether or not event A has occurred, then event A does not affect event B and

P(B͉A) ϭ P(B)

The Multiplication Rule can now be simpliﬁed.

THE MULTIPLICATION RULE FOR INDEPENDENT

EVENTS

If two events A and B are independent, the probability that both A and B occur is

P(A ʝ B) ϭ P(A)P(B)

Similarly, if A, B, and C are mutually independent events (all pairs of events are

independent), then the probability that A, B, and C all occur is

P(A ʝ B ʝ C) ϭ P(A)P(B)P(C)

152

CHAPTER 4 PROBABILITY AND PROBABILITY DISTRIBUTIONS

Coin Tosses at Football Games A football team is involved in two overtime periods during a given game, so that there are three coins tosses. If the coin is

fair, what is the probability that they lose all three tosses?

Solution

If the coin is fair, the event can be described in three steps:

A: lose the ﬁrst toss

B: lose the second toss

C: lose the third toss

Since the tosses are independent, and since P(win) ϭ P(lose) ϭ .5 for any of the three

tosses,

P(A ʝ B ʝ C) ϭ P(A)P(B)P(C) ϭ (.5)(.5)(.5) ϭ .125

How can you check to see if two events are independent or dependent? The easiest

solution is to redeﬁne the concept of independence in a more formal way.

CHECKING FOR INDEPENDENCE

Two events A and B are said to be independent if and only if either

P(A ʝ B) ϭ P(A)P(B)

or

P(B͉A) ϭ P(B)

Otherwise, the events are said to be dependent.

EXAMPLE

4.20

Toss two coins and observe the outcome. Deﬁne these events:

A: Head on the ﬁrst coin

B: Tail on the second coin

Are events A and B independent?

From previous examples, you know that S ϭ {HH, HT, TH, TT}. Use

these four simple events to ﬁnd

Solution

Remember,

independence ⇔

P(A ʝ B) ϭ P (A)P(B).

1

1

1

P(A) ϭ ᎏᎏ, P(B) ϭ ᎏᎏ, and P(A ʝ B) ϭ ᎏᎏ.

2

2

4

΂ ΃΂ ΃

1 1

1

1

Since P(A)P(B) ϭ ᎏᎏ ᎏᎏ ϭ ᎏᎏ and P(A ʝ B) ϭ ᎏᎏ, we have P(A)P(B) ϭ P(A ʝ B)

2 2

4

4

and the two events must be independent.

EXAMPLE

4.21

Refer to the probability table in Example 4.18, which is reproduced below.

Child in College (D)

No Child in College (E )

Too High

(A)

Right Amount

(B)

Too Little

(C )

.35

.25

.08

.20

.01

.11

Are events D and A independent? Explain.

4.6 INDEPENDENCE, CONDITIONAL PROBABILITY, AND THE MULTIPLICATION RULE

153

Solution

1. Use the probability table to ﬁnd P(A ʝ D) ϭ .35, P(A) ϭ .60, and

P(D) ϭ .44. Then

P(A)P(D) ϭ (.60)(.44) ϭ .264 and P(A ʝ D) ϭ .35

Since these two probabilities are not the same, events A and D are dependent.

2. Alternately, calculate

P(A ʝ D)

.35

P(A͉ D) ϭ ᎏᎏ ϭ ᎏᎏ ϭ .80

P(D)

.44

Since P(A͉ D) ϭ .80 and P(A) ϭ .60, we are again led to the conclusion that

events A and D are dependent.

What’s the Difference between Mutually Exclusive

and Independent Events?

Many students ﬁnd it hard to tell the difference between mutually exclusive and

independent events.

• When two events are mutually exclusive or disjoint, they cannot both happen when the experiment is performed. Once the event B has occurred,

event A cannot occur, so that P(A͉B) ϭ 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.

When two events are mutually exclusive or disjoint,

P(A ʝ B) ϭ 0 and P(A ʜ B) ϭ P(A) ϩ P(B).

When two events are independent,

P(A ʝ B) ϭ P(A)P(B), and P(A ʜ B) ϭ P(A) ϩ P(B) Ϫ P(A)P(B).

Exercise Reps

Use the relationships above to ﬁll in the blanks in the table below.

P(A)

P(B)

.3

.4

Conditions for Events A and B

Mutually exclusive

.3

.4

Independent

.1

.5

.2

.5

P(A 8 B)

P(A 9 B)

P(AͦB)

.6

.10

Answers are located on the perforated card at the back of this book.

Using probability rules to calculate the probability of an event requires some experience and ingenuity. You need to express the event of interest as a union or intersection (or the combination of both) of two or more events whose probabilities are

known or easily calculated. Often you can do this in different ways; the key is to ﬁnd

the right combination.

154

EXAMPLE

CHAPTER 4 PROBABILITY AND PROBABILITY DISTRIBUTIONS

4.22

Two cards are drawn from a deck of 52 cards. Calculate the probability that the draw

includes an ace and a ten.

Solution

Consider the event of interest:

A: Draw an ace and a ten

Then A ϭ B ʜ C, where

B: Draw the ace on the ﬁrst draw and the ten on the second

C: Draw the ten on the ﬁrst draw and the ace on the second

Events B and C were chosen to be mutually exclusive and also to be intersections of

events with known probabilities; that is,

B ϭ B1 ʝ B2 and C ϭ C1 ʝ C2

where

B1: Draw an ace on the ﬁrst draw

B2: Draw a ten on the second draw

C1: Draw a ten on the ﬁrst draw

C2: Draw an ace on the second draw

Applying the Multiplication Rule, you get

P(B1 ʝ B2) ϭ P(B1)P(B2͉B1)

΂ ΃΂ ΃

4

4

ϭ ᎏᎏ ᎏᎏ

52 51

and

΂ ΃΂ ΃

4

4

P(C1 ʝ C2) ϭ ᎏᎏ ᎏᎏ

52 51

Then, applying the Addition Rule,

P(A) ϭ P(B) ϩ P(C)

΂ ΃΂ ΃ ΂ ΃΂ ΃

4

4

4

4

8

ϭ ᎏᎏ ᎏᎏ ϩ ᎏᎏ ᎏᎏ ϭ ᎏᎏ

52 51

52 51

663

Check each composition carefully to be certain that it is actually equal to the event of

interest.

4.6

EXERCISES

EXERCISE REPS

These exercises refer to the MyPersonal Trainer section on page 153.

4.40 Use event relationships to ﬁll in the blanks in the table below.

P(A)

P(B)

.3

.4

Conditions for Events A and B

.3

.4

.1

.5

Mutually exclusive

.2

.5

Independent

P(A 8 B)

P(A 9 B)

.12

.7

P(AͦB)

4.6 INDEPENDENCE, CONDITIONAL PROBABILITY, AND THE MULTIPLICATION RULE

155

4.41 Use event relationships to ﬁll in the blanks in the table below.

P(A)

P(B)

Conditions for Events A and B

.3

.4

Mutually exclusive

.3

.4

Independent

.1

.5

.2

.5

P(A 8 B)

P(A 9 B)

P(AͦB)

.1

0

BASIC TECHNIQUES

A: Observe a number less than 4

4.42 An experiment can result in one of ﬁve equally

B: Observe a number less than or equal to 2

C: Observe a number greater than 3

likely simple events, E1, E2, . . . , E5. Events A, B, and

C are deﬁned as follows:

A: E1, E3

P(A) ϭ .4

B: E1, E2, E4, E5

C: E3, E4

P(B) ϭ .8

P(C) ϭ .4

Find the probabilities associated with these compound

events by listing the simple events in each.

b. A ʝ B

c. B ʝ C

a. Ac

d. A ʜ B

e. B͉C

f. A͉B

g. A ʜ B ʜ C

h. (A ʝ B)c

4.43 Refer to Exercise 4.42. Use the deﬁnition of a

complementary event to ﬁnd these probabilities:

a. P(Ac )

b. P((A ʝ B)c )

Do the results agree with those obtained in Exercise 4.42?

4.44 Refer to Exercise 4.42. Use the deﬁnition of

conditional probability to ﬁnd these probabilities:

a. P(A͉B)

b. P(B͉C)

Do the results agree with those obtained in Exercise 4.42?

4.45 Refer to Exercise 4.42. Use the Addition and

Multiplication Rules to ﬁnd these probabilities:

a. P(A ʜ B)

b. P(A ʝ B)

c. P(B ʝ C)

Do the results agree with those obtained in Exercise 4.42?

Find the probabilities associated with the events below

using either the simple event approach or the rules and

deﬁnitions from this section.

a. S

b. A͉B

c. B

d. A ʝ B ʝ C e. A ʝ B

f. A ʝ C

g. B ʝ C

h. A ʜ C

i. B ʜ C

4.48 Refer to Exercise 4.47.

a. Are events A and B independent? Mutually exclusive?

b. Are events A and C independent? Mutually exclusive?

4.49 Suppose that P(A) ϭ .4 and P(B) ϭ .2. If events

A and B are independent, ﬁnd these probabilities:

a. P(A ʝ B)

b. P(A ʜ B)

4.50 Suppose that P(A) ϭ .3 and P(B) ϭ .5. If

events A and B are mutually exclusive, ﬁnd these

probabilities:

a. P(A ʝ B)

b. P(A ʜ B)

4.51 Suppose that P(A) ϭ .4 and P(A ʝ B) ϭ .12.

a. Find P(B͉A).

b. Are events A and B mutually exclusive?

c. If P(B) ϭ .3, are events A and B independent?

4.52 An experiment can result in one or both of

events A and B with the probabilities shown in this

probability table:

4.46 Refer to Exercise 4.42.

a. Are events A and B independent?

b. Are events A and B mutually exclusive?

4.47 Dice An experiment consists of tossing a single

die and observing the number of dots that show on the

upper face. Events A, B, and C are deﬁned as follows:

B

Bc

A

Ac

.34

.15

.46

.05

Find the following probabilities:

a. P(A)

d. P(A ʜ B)

b. P(B)

e. P(A͉B)

c. P(A ʝ B)

f. P(B͉A)

156

CHAPTER 4 PROBABILITY AND PROBABILITY DISTRIBUTIONS

4.53 Refer to Exercise 4.52.

a. Are events A and B mutually exclusive? Explain.

b. Are events A and B independent? Explain.

A: The offender has 10 or more years of education

B: The offender is convicted within 2 years after

completion of treatment

APPLICATIONS

4.54 Drug Testing Many companies are testing

prospective employees for drug use, with the intent of

improving efficiency and reducing absenteeism, accidents, and theft. Opponents claim that this procedure is

creating a class of unhirables and that some persons

may be placed in this class because the tests themselves are not 100% reliable. Suppose a company uses

a test that is 98% accurate—that is, it correctly identiﬁes a person as a drug user or nonuser with probability

.98—and to reduce the chance of error, each job applicant is required to take two tests. If the outcomes of

the two tests on the same person are independent

events, what are the probabilities of these events?

a. A nonuser fails both tests.

b. A drug user is detected (i.e., he or she fails at least

one test).

c. A drug user passes both tests.

4.55 Grant Funding Whether a grant proposal is

funded quite often depends on the reviewers. Suppose

a group of research proposals was evaluated by a

group of experts as to whether the proposals were worthy of funding. When these same proposals were submitted to a second independent group of experts, the

decision to fund was reversed in 30% of the cases. If

the probability that a proposal is judged worthy of

funding by the ﬁrst peer review group is .2, what are

the probabilities of these events?

a. A worthy proposal is approved by both groups.

b. A worthy proposal is disapproved by both groups.

c. A worthy proposal is approved by one group.

4.56 Drug Offenders A study of the behavior of a

large number of drug offenders after treatment for drug

abuse suggests that the likelihood of conviction within

a 2-year period after treatment may depend on the

offender’s education. The proportions of the total

number of cases that fall into four education/conviction

categories are shown in the table below.

Status Within 2

Years After Treatment

Convicted

Not

Convicted

Totals

10 Years or More

9 Years or Less

.10

.27

.30

.33

.40

.60

Totals

.37

.63

1.00

Education

Suppose a single offender is selected from the treatment program. Here are the events of interest:

Find the appropriate probabilities for these events:

a. A

b. B

c. A ʝ B

d. A ʜ B

e. Ac

f. (A ʜ B)c

g. (A ʝ B)c

h. A given that B has occurred

i. B given that A has occurred

4.57 Use the probabilities of Exercise 4.56 to show

that these equalities are true:

a. P(A ʝ B) ϭ P(A)P(B͉A)

b. P(A ʝ B) ϭ P(B)P(A͉B)

c. P(A ʜ B) ϭ P(A) ϩ P(B) Ϫ P(A ʝ B)

4.58 The Birthday Problem Two people enter a

room and their birthdays (ignoring years) are recorded.

a. Identify the nature of the simple events in S.

b. What is the probability that the two people have a

speciﬁc pair of birthdates?

c. Identify the simple events in event A: Both people

have the same birthday.

d. Find P(A).

e. Find P(Ac).

4.59 The Birthday Problem, continued If n peo-

ple enter a room, ﬁnd these probabilities:

A: None of the people have the same birthday

B: At least two of the people have the same birthday

Solve for

a. n ϭ 3

b. n ϭ 4

[NOTE: Surprisingly, P(B) increases rapidly as n

increases. For example, for n ϭ 20, P(B) ϭ .411;

for n ϭ 40, P(B) ϭ .891.]

4.60 Starbucks or Peet’s®? A college student fre-

quents one of two coffee houses on campus, choosing

Starbucks 70% of the time and Peet’s 30% of the time.

Regardless of where she goes, she buys a cafe mocha

on 60% of her visits.

a. The next time she goes into a coffee house on campus, what is the probability that she goes to Starbucks and orders a cafe mocha?

b. Are the two events in part a independent? Explain.

c. If she goes into a coffee house and orders a cafe

mocha, what is the probability that she is at Peet’s?

4.6 INDEPENDENCE, CONDITIONAL PROBABILITY, AND THE MULTIPLICATION RULE

d. What is the probability that she goes to Starbucks

or orders a cafe mocha or both?

4.61 Inspection Lines A certain manufactured item

is visually inspected by two different inspectors. When

a defective item comes through the line, the probability that it gets by the ﬁrst inspector is .1. Of those that

get past the ﬁrst inspector, the second inspector will

“miss” 5 out of 10. What fraction of the defective

items get by both inspectors?

4.62 Smoking and Cancer A survey of people in a

given region showed that 20% were smokers. The

probability of death due to lung cancer, given that a

person smoked, was roughly 10 times the probability

of death due to lung cancer, given that a person did not

smoke. If the probability of death due to lung cancer in

the region is .006, what is the probability of death due

to lung cancer given that a person is a smoker?

4.63 Smoke Detectors A smoke-detector system

uses two devices, A and B. If smoke is present, the

probability that it will be detected by device A is .95;

by device B, .98; and by both devices, .94.

a. If smoke is present, ﬁnd the probability that the

smoke will be detected by device A or device B or

both devices.

b. Find the probability that the smoke will not be

detected.

4.64 Plant Genetics Gregor Mendel was a monk

who suggested in 1865 a theory of inheritance based

on the science of genetics. He identiﬁed heterozygous

individuals for ﬂower color that had two alleles (one

r ϭ recessive white color allele and one R ϭ dominant

red color allele). When these individuals were mated,

3/4 of the offspring were observed to have red ﬂowers

and 1/4 had white ﬂowers. The table summarizes this

mating; each parent gives one of its alleles to form the

gene of the offspring.

Parent 2

Parent 1

r

R

r

R

rr

Rr

rR

RR

We assume that each parent is equally likely to give

either of the two alleles and that, if either one or two

of the alleles in a pair is dominant (R), the offspring

will have red ﬂowers.

a. What is the probability that an offspring in this mating has at least one dominant allele?

b. What is the probability that an offspring has at least

one recessive allele?

157

c. What is the probability that an offspring has one

recessive allele, given that the offspring has red

ﬂowers?

4.65 Soccer Injuries During the inaugural season of

Major League Soccer in the United States, the medical

teams documented 256 injuries that caused a loss of

participation time to the player. The results of this

investigation, reported in The American Journal of

Sports Medicine, is shown in the table.3

Severity

Minor (A)

Moderate (B)

Major (C)

Total

Practice (P)

Game (G)

Total

66

23

12

88

44

23

154

67

35

101

155

256

If one individual is drawn at random from this group

of 256 soccer players, ﬁnd the following probabilities:

a. P(A)

b. P(G)

c. P(A ʝ G)

d. P(G͉A)

e. P(G͉B)

f. P(G͉C)

g. P(C͉P)

h. P(Bc )

4.66 Choosing a Mate Men and women often disagree on how they think about selecting a mate. Suppose that a poll of 1000 individuals in their twenties

gave the following responses to the question of

whether it is more important for their future mate to be

able to communicate their feelings (F ) than it is for

that person to make a good living (G).

Feelings (F )

Good Living (G)

Totals

Men (M)

Women (W )

.35

.36

.20

.09

.55

.45

Totals

.71

.29

1.00

If an individual is selected at random from this

group of 1000 individuals, calculate the following

probabilities:

a. P(F)

d. P(F͉W)

b. P(G)

e. P(M͉F)

c. P(F͉M)

f. P(W͉G)

4.67 Jason and Shaq The two stars of the Miami

Heat professional basketball team are very different

when it comes to making free throws. ESPN.com

reports that Jason Williams makes about 80% of his

free throws, while Shaquille O’Neal makes only 53%

of his free throws.4 Assume that the free throws are

independent, and that each player takes two free

throws during a particular game.

a. What is the probability that Jason makes both of

his free throws?

158 ❍

CHAPTER 4 PROBABILITY AND PROBABILITY DISTRIBUTIONS

b. What is the probability that Shaq makes exactly

Player A has probability 1/6 of winning the tournament

if player B enters and probability 3/4 of winning if

player B does not enter the tournament. If the probability that player B enters is 1/3, ﬁnd the probability that

player A wins the tournament.

one of his two free throws?

c. What is the probability that Shaq makes both of

his free throws and Jason makes neither of his?

4.68 Golﬁng Player A has entered a golf tournament

but it is not certain whether player B will enter.

BAYES’ RULE (OPTIONAL)

4.7

Let us reconsider the experiment involving colorblindness from

Section 4.6. Notice that the two events

Colorblindness

B: the person selected is a man

BC: the person selected is a woman

taken together make up the sample space S, consisting of both men and women. Since

colorblind people can be either male or female, the event A, which is that a person is

colorblind, consists of both those simple events that are in A and B and those simple

events that are in A and BC. Since these two intersections are mutually exclusive, you

can write the event A as

A ϭ (A ʝ B) ʜ (A ʝ BC)

and

P(A) ϭ P(A ʝ B) ϩ P(A ʝ BC)

ϭ .04 ϩ .002 ϭ .042

Suppose now that the sample space can be partitioned into k subpopulations,

S1, S2, S3, . . . , Sk, that, as in the colorblindness example, are mutually exclusive and

exhaustive; that is, taken together they make up the entire sample space. In a similar

way, you can express an event A as

A ϭ (A ʝ S1) ʜ (A ʝ S2 ) ʜ (A ʝ S3 )ʜ и и и ʜ (A ʝ Sk )

Then

P(A) ϭ P(A ʝ S1) ϩ P(A ʝ S2 ) ϩ P(A ʝ S3 ) ϩ и и и ϩ P(A ʝ Sk )

This is illustrated for k ϭ 3 in Figure 4.14.

F IGU R E 4 .1 4

Decomposition of event A

S

A∩S1

S1

A∩S2

S2

A∩S3

S3

4.7 BAYES’ RULE (OPTIONAL)

159

You can go one step further and use the Multiplication Rule to write P(A ʝ Si) as

P(Si)P(A͉Si), for i ϭ 1, 2, . . . , k. The result is known as the Law of Total Probability.

LAW OF TOTAL PROBABILITY

Given a set of events S1, S2, S3, . . . , Sk that are mutually exclusive and exhaustive and an event A, the probability of the event A can be expressed as

P(A) ϭ P(S1)P(A͉S1) ϩ P(S2)P(A͉S2) ϩ P(S3)P(A͉S3) ϩ и и и ϩ P(Sk)P(A͉Sk)

EXAMPLE

TABLE 4.6

4.23

Sneakers are no longer just for the young. In fact, most adults own multiple pairs of

sneakers. Table 4.6 gives the fraction of U.S. adults 20 years of age and older who

own ﬁve or more pairs of wearable sneakers, along with the fraction of the U.S. adult

population 20 years or older in each of ﬁve age groups.5 Use the Law of Total Probability to determine the unconditional probability of an adult 20 years and older owning ﬁve or more pairs of wearable sneakers.

Probability Table

Groups and Ages

Fraction with Ն 5 Pairs

Fraction of U.S. Adults 20 and Older

G1

20–24

G2

25–34

G3

35–49

G4

50–64

G5

Ն 65

.26

.09

.20

.20

.13

.31

.18

.23

.14

.17

Let A be the event that a person chosen at random from the U.S. adult

population 18 years of age and older owns ﬁve or more pairs of wearable sneakers.

Let G1, G2, . . . , G5 represent the event that the person selected belongs to each of

the ﬁve age groups, respectively. Since the ﬁve groups are exhaustive, you can write

the event A as

Solution

A ϭ (A ʝ G1) ʜ (A ʝ G2) ʜ (A ʝ G3) ʜ (A ʝ G4) ʜ (A ʝ G5)

Using the Law of Total Probability, you can ﬁnd the probability of A as

P(A) ϭ P(A ʝ G1) ϩ P(A ʝ G2) ϩ P(A ʝ G3) ϩ P(A ʝ G4) ϩ P(A ʝ G5)

ϭ P(G1)P(A͉G1) ϩ P(G2)P(A͉G2) ϩ P(G3)P(A͉G3)

ϩ P(G4)P(A͉G4) ϩ P(G5)P(A͉G5)

From the probabilities in Table 4.6,

P(A) ϭ (.09)(.26) ϩ (.20)(.20) ϩ (.31)(.13) ϩ (.23)(.18) ϩ (.17)(.14)

ϭ .0234 ϩ .0400 ϩ .0403 ϩ .0414 ϩ .0238 ϭ .1689

The unconditional probability that a person selected at random from the population

of U.S. adults 20 years of age and older owns at least ﬁve pairs of wearable sneakers

is about .17. Notice that the Law of Total Probability is a weighted average of the

probabilities within each group, with weights .09, .20, .31, .23, and .17, which reﬂect

the relative sizes of the groups.

Often you need to ﬁnd the conditional probability of an event B, given that an

event A has occurred. One such situation occurs in screening tests, which used to be

160

CHAPTER 4 PROBABILITY AND PROBABILITY DISTRIBUTIONS

associated primarily with medical diagnostic tests but are now ﬁnding applications in

a variety of ﬁelds. Automatic test equipment is routinely used to inspect parts in highvolume production processes. Steroid testing of athletes, home pregnancy tests,

and AIDS testing are some other applications. Screening tests are evaluated on the

probability of a false negative or a false positive, and both of these are conditional

probabilities.

A false positive is the event that the test is positive for a given condition, given

that the person does not have the condition. A false negative is the event that the test

is negative for a given condition, given that the person has the condition. You can

evaluate these conditional probabilities using a formula derived by the probabilist

Thomas Bayes.

The experiment involves selecting a sample from one of k subpopulations that

are mutually exclusive and exhaustive. Each of these subpopulations, denoted by

S1, S2, . . . , Sk, has a selection probability P(S1), P(S2), P(S3), . . . , P(Sk), called prior

probabilities. An event A is observed in the selection. What is the probability that the

sample came from subpopulation Si, given that A has occurred?

You know from Section 4.6 that P(Si |A) ϭ [P(A ʝ Si)]/P(A), which can be rewritten as P(Si |A) ϭ [P(Si)P(A|Si)]/P(A). Using the Law of Total Probability to rewrite

P(A), you have

P(Si)P(A|Si)

P(Si |A) ϭ ᎏᎏᎏᎏᎏᎏᎏ

P(S1)P(A|S1) ϩ P(S2)P(A|S2) ϩ P(S3)P(A|S3) ϩ и и и ϩ P(Sk)P(A|Sk)

These new probabilities are often referred to as posterior probabilities—that is, probabilities of the subpopulations (also called states of nature) that have been updated

after observing the sample information contained in the event A. Bayes suggested that

if the prior probabilities are unknown, they can be taken to be 1/k, which implies that

each of the events S1 through Sk is equally likely.

BAYES’ RULE

Let S1, S2, . . . , Sk represent k mutually exclusive and exhaustive subpopulations

with prior probabilities P(S1), P(S2), . . . , P(Sk). If an event A occurs, the posterior probability of Si given A is the conditional probability

P(Si)P(A|Si)

P(Si |A) ϭ ᎏᎏ

k

Α P(Sj)P(A|Sj)

jϭ1

for i ϭ 1, 2, . . . , k.

EXAMPLE

4.24

Refer to Example 4.23. Find the probability that the person selected was 65 years of

age or older, given that the person owned at least ﬁve pairs of wearable sneakers.

Solution

You need to ﬁnd the conditional probability given by

P(A ʝ G5)

P(G5|A) ϭ ᎏᎏ

P(A)

You have already calculated P(A) ϭ .1689 using the Law of Total Probability.

Therefore,

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