76197039 Ken Black QA 5th Chapter 4 Solution

Chapter 4: Probability 1

Chapter 4Probability

LEARNING OBJECTIVES

The main objective of Chapter 4 is to help you understand the basic principles of probability, specifically enabling you to

1. Comprehend the different ways of assigning probability.2. Understand and apply marginal, union, joint, and conditional probabilities. 3. Select the appropriate law of probability to use in solving problems.4. Solve problems using the laws of probability, including the law of addition, the

law of multiplication , and the law of conditional probability.5. Revise probabilities using Bayes' rule.

CHAPTER TEACHING STRATEGY

Students can be motivated to study probability by realizing that the field of probability has some stand-alone application in their lives in such applied areas human resource analysis, actuarial science, and gaming. In addition, students should understand that much of the rest of this course is based on probabilities even though they will not be directly applying many of these formulas in other chapters.

This chapter is frustrating for the learner because probability problems can be approached by using several different techniques. Whereas, in many chapters of this text, students will approach problems by using one standard technique, in chapter 4, different students will often use different approaches to the same problem. The text attempts to emphasize this point and underscore it by presenting several different ways to solve probability problems. The probability rules and laws presented in the chapter can virtually always be used in solving probability problems. However, it is sometimes easier to construct a probability matrix or a tree diagram or use the sample space to solve the problem. If the student is aware that what they have at their hands is an array of tools or techniques, they will be less overwhelmed in approaching a probability problem. An attempt has been made to differentiate the several types of probabilities so that students can sort out the various types of problems.


In teaching students how to construct a probability matrix, emphasize that it is usually best to place only one variable along each of the two dimensions of the matrix. (That is, place Mastercard with yes/no on one axis and Visa with yes/no on the other instead of trying to place Mastercard and Visa along the same axis).

This particular chapter is very amenable to the use of visual aids. Students enjoy rolling dice, tossing coins, and drawing cards as a part of the class experience. Of all the chapters in the book, it is most imperative that students work a lot of problems in this chapter. Probability problems are so varied and individualized that a significant portion of the learning comes in the doing. Experience is an important factor in working probability problems.

Section 4.8 on Bayes’ theorem can be skipped in a one-semester course without losing any continuity. This section is a prerequisite to the chapter 19 presentation of “revising probabilities in light of sample information (section 19.4).

CHAPTER OUTLINE

4.1 Introduction to Probability

4.2 Methods of Assigning ProbabilitiesClassical Method of Assigning ProbabilitiesRelative Frequency of OccurrenceSubjective Probability

4.3 Structure of ProbabilityExperimentEventElementary EventsSample SpaceUnions and IntersectionsMutually Exclusive EventsIndependent EventsCollectively Exhaustive EventsComplimentary EventsCounting the Possibilities

The mn Counting Rule Sampling from a Population with Replacement

Combinations: Sampling from a Population Without Replacement

4.4 Marginal, Union, Joint, and Conditional Probabilities

4.5 Addition LawsProbability Matrices

Complement of a UnionSpecial Law of Addition


4.6 Multiplication LawsGeneral Law of MultiplicationSpecial Law of Multiplication

4.7 Conditional ProbabilityIndependent Events

4.8 Revision of Probabilities: Bayes' Rule

KEY TERMS

A Priori IntersectionBayes' Rule Joint ProbabilityClassical Method of Assigning Probabilities Marginal ProbabilityCollectively Exhaustive Events mn Counting RuleCombinations Mutually Exclusive EventsComplement of a Union Probability MatrixComplement Relative Frequency of OccurrenceConditional Probability Sample SpaceElementary Events Set NotationEvent Subjective ProbabilityExperiment UnionIndependent Events Union Probability


SOLUTIONS TO PROBLEMS IN CHAPTER 4

4.1 Enumeration of the six parts: D1, D2, D3, A4, A5, A6

D = Defective partA = Acceptable part

Sample Space:

D1 D2, D2 D3, D3 A5

D1 D3, D2 A4, D3 A6

D1 A4, D2 A5, A4 A5

D1 A5, D2 A6, A4 A6

D1 A6, D3 A4, A5 A6

There are 15 members of the sample space

The probability of selecting exactly one defect out of two is:

9/15 = .60

4.2 X = {1, 3, 5, 7, 8, 9}, Y = {2, 4, 7, 9}, and Z = {1, 2, 3, 4, 7,} a) X ⊥ Z = {1, 2, 3, 4, 5, 7, 8, 9}

b) X _ Y = {7, 9} c) X _ Z = {1, 3, 7} d) X ⊥ Y ⊥ Z = {1, 2, 3, 4, 5, 7, 8, 9} e) X _ Y _ Z = {7} f) (X ⊥ Y) _ Z = {1, 2, 3, 4, 5, 7, 8, 9} _ {1, 2, 3, 4, 7} = {1, 2, 3, 4, 7} g) (Y _ Z) ⊥ (X _ Y) = {2, 4, 7} ⊥ {7, 9} = {2, 4, 7, 9}

h) X or Y = X ⊥ Y = {1, 2, 3, 4, 5, 7, 8, 9}i) Y and Z = Y _ Z = {2, 4, 7}


4.3 If A = {2, 6, 12, 24} and the population is the positive even numbers through 30,

A’ = {4, 8, 10, 14, 16, 18, 20, 22, 26, 28, 30}

4.4 6(4)(3)(3) = 216

4.5 Enumeration of the six parts: D1, D2, A1, A2, A3, A4

D = Defective partA = Acceptable part

Sample Space:

D1 D2 A1, D1 D2 A2, D1 D2 A3,D1 D2 A4, D1 A1 A2, D1 A1 A3,D1 A1 A4, D1 A2 A3, D1 A2 A4,D1 A3 A4, D2 A1 A2, D2 A1 A3,D2 A1 A4, D2 A2 A3, D2 A2 A4,D2 A3 A4, A1 A2 A3, A1 A2 A4,A1 A3 A4, A2 A3 A4

Combinations are used to counting the sample space because sampling is done without replacement.

6C3 = !3!3

!6 = 20

Probability that one of three is defective is:

12/20 = 3/5 .60

There are 20 members of the sample space and 12 of them have exactly 1 defective part.

4.6 107 = 10,000,000 different numbers

4.7 20C6 = !14!6

!20 = 38,760

It is assumed here that 6 different (without replacement) employees are to be selected.


4.8 P(A) = .10, P(B) = .12, P(C) = .21, P(A _ C) = .05 P(B _ C) = .03

a) P(A ⊥ C) = P(A) + P(C) - P(A _ C) = .10 + .21 - .05 = .26

b) P(B ⊥ C) = P(B) + P(C) - P(B _ C) = .12 + .21 - .03 = .30

c) If A, B mutually exclusive, P(A ⊥ B) = P(A) + P(B) = .10 + .12 = .22

4.9 D E F

A 5 8 12 25

B 10 6 4 20

C 8 2 5 15

23 16 21 60

a) P(A ⊥ D) = P(A) + P(D) - P(A _ D) = 25/60 + 23/60 - 5/60 = 43/60 = .7167

b) P(E ⊥ B) = P(E) + P(B) - P(E _ B) = 16/60 + 20/60 - 6/60 = 30/60 = .5000c) P(D ⊥ E) = P(D) + P(E) = 23/60 + 16/60 = 39/60 = .6500d) P(C ⊥ F) = P(C) + P(F) - P(C _ F) = 15/60 + 21/60 - 5/60 = 31/60 = .

5167


4.10

E F

A .10 .03 .13

B .04 .12 .16

C .27 .06 .33

D .31 .07 .38

.72 .28 1.00

a) P(A ⊥ F) = P(A) + P(F) - P(A _ F) = .13 + .28 - .03 = .38

b) P(E ⊥ B) = P(E) + P(B) - P(E _ B) = .72 + .16 - .04 = .84 c) P(B ⊥ C) = P(B) + P(C) =.16 + .33 = .49 d) P(E ⊥ F) = P(E) + P(F) = .72 + .28 = 1.000

4.11 A = event of having flown in an airplane at least onceT = event of having ridden in a train at least once

P(A) = .47 P(T) = .28

P (ridden either a train or an airplane) =P(A ⊥ T) = P(A) + P(T) - P(A _ T) = .47 + .28 - P(A _ T)

Cannot solve this problem without knowing the probability of the intersection. We need to know the probability of the intersection of A and T, the proportion who have ridden both or determine if these two events are mutually exclusive.

4.12 P(L) = .75 P(M) = .78 P(M _ L) = .61a) P(M ⊥ L) = P(M) + P(L) - P(M _ L) = .78 + .75 - .61 = .92 b) P(M ⊥ L) but not both = P(M ⊥ L) - P(M _ L) = .92 - .61 = .31

c) P(NM _ NL) = 1 - P(M ⊥ L) = 1 - .92 = .08 Note: the neither/nor event is solved for here by taking the complement of the

union.


4.13 Let C = have cable TV Let T = have 2 or more TV sets P(C) = .67, P(T) = .74, P(C _ T) = .55

a) P(C ⊥ T) = P(C) + P(T) - P(C _ T) = .67 + .74 - .55 = .86b) P(C ⊥ T but not both) = P(C ⊥ T) - P(C _ T) = .86 - .55 = .31c) P(NC _ NT) = 1 - P(C ⊥ T) = 1 - .86 = .14d) The special law of addition does not apply because P(C _ T) is not .0000. Possession of cable TV and 2 or more TV sets are not mutually exclusive.

4.14 Let T = review transcript F = consider faculty referencesP(T) = .54 P(F) = .44 P(T _ F) = .35 a) P(F ⊥ T) = P(F) + P(T) - P(F _ T) = .44 + .54 - .35 = .63

b) P(F ⊥ T) - P(F _ T) = .63 - .35 = .28 c) 1 - P(F ⊥ T) = 1 - .63 = .37

d) Faculty References Y N

Transcript Y .35 .19 .54

N .09 .37 .46

.44 .56 1.00


4.15 C D E F

A 5 11 16 8 40

B 2 3 5 7 17

7 14 21 15 57

a) P(A _ E) = 16/57 = .2807 b) P(D _ B) = 3/57 = .0526 c) P(D _ E) = .0000 d) P(A _ B) = .0000

4.16 D E F

A .12 .13 .08 .33

B .18 .09 .04 .31

C .06 .24 .06 .36

.36 .46 .18 1.00

a) P(E _ B) = .09 b) P(C _ F) = .06 c) P(E _ D) = .00

4.17 Let D = Defective part

a) (without replacement)

P(D1 _ D2) = P(D1) ⋅ P(D2 D1) = 2450

30

49

5

50

6 =⋅ = .0122

b) (with replacement)

P(D1 _ D2) = P(D1) ⋅ P(D2) = 2500

36

50

6

50

6 =⋅ = .0144

4.18 Let U = Urban I = care for Ill relatives

P(U) = .78 P(I) = .15 P(IU) = .11 a) P(U _ I) = P(U) ⋅ P(IU)

P(U _ I) = (.78)(.11) = .0858


b) P(U _ NI) = P(U) ⋅ P(NIU) but P(IU) = .11 So, P(NIU) = 1 - .11 = .89 and

P(U _ NI) = P(U) ⋅ P(NIU) = (.78)(.89) = .6942

c) U

Yes No

I Yes .15

No .85

.78 .22

The answer to a) is found in the YES-YES cell. To compute this cell, take 11% or .11 of the total (.78) people in urban areas. (.11)(.78) = .0858 which belongs in the “YES-YES" cell. The answer to b) is found in the Yes for U and no for I cell. It can be determined by taking the marginal, .78, less the answer for a), .0858. d. P(NU _ I) is found in the no for U column and the yes for I row (1st row and 2nd column). Take the marginal, .15, minus the yes-yes cell, .0858, to get .0642.


4.19 Let S = stockholder Let C = college

P(S) = .43 P(C) = .37 P(CS) = .75

a) P(NS) = 1 - .43 = .57 b) P(S _ C) = P(S)⋅ P(CS) = (.43)(.75) = .3225 c) P(S ⊥ C) = P(S) + P(C) - P(S _ C) = .43 + .37 - .3225 = .4775 d) P(NS _ NC) = 1 - P(S ⊥ C) = 1 - .4775 = .5225

e) P(NS ⊥ NC) = P(NS) + P(NC) - P(NS _ NC) = .57 + .63 - .5225 = .6775 f) P(C _ NS) = P(C) - P(C _ S) = .37 - .3225 = .0475

The matrix:

C

Yes No

S Yes .3225 .1075 .43

No .0475 .5225 .57

.37 .63 1.00


4.20 Let F = fax machine Let P = personal computer

Given: P(F) = .10 P(P) = .52 P(PF) = .91a) P(F _ P) = P(F) ⋅ P(P F) = (.10)(.91) = .091b) P(F ⊥ P) = P(F) + P(P) - P(F _ P) = .10 + .52 - .091 = .529c) P(F _ NP) = P(F) ⋅ P(NP F)

Since P(P F) = .91, P(NP F)= 1 - P(P F) = 1 - .91 = .09 P(F _ NP) = (.10)(.09) = .009

d) P(NF _ NP) = 1 - P(F ⊥ P) = 1 - .529 = .471e) P(NF _ P) = P(P) - P(F _ P) = .52 - .091 = .429

The matrix:

P NP

F .091 .009 .10

NF .429 .471 .90

.520 .480 1.00


4.21Let S = safety Let A = age

P(S) = .30 P(A) = .39 P(AS) = .87a) P(S _ NA) = P(S)⋅ P(NAS)

but P(NAS) = 1 - P(AS) = 1 - .87 = .13 P(S _ NA) = (.30)(.13) = .039

b) P(NS _ NA) = 1 - P(S ⊥ A) = 1 - [P(S) + P(A) - P(S _ A)] but P(S _ A) = P(S) ⋅ P(AS) = (.30)(.87) = .261 P(NS _ NA) = 1 - (.30 + .39 - .261) = .571

c) P(NS _ A) = P(NS) - P(NS _ NA)

but P(NS) = 1 - P(S) = 1 - .30 = .70 P(NS _ A) = .70 - 571 = .129

The matrix:

A

Yes No

S Yes .261 .039 .30

No .129 .571 .70

.39 .61 1.00


4.22 Let C = ceiling fans Let O = outdoor grillP(C) = .60 P(O) = .29 P(C _ O) = .13a) P(C ⊥ O)= P(C) + P(O) - P(C _ O) = .60 + .29 - .13 = .76b) P(NC _ NO) = 1 - P(C ⊥ O)= 1 - .76 = .24c) P(NC _ O) = P(O) - P(C _ O) = .29 - .13 = .16d) P(C _ NO) = P(C) - P(C _ O) = .60 - .13 = .47

The matrix:

O

Yes No

C Yes .13 .47 .60

No .16 .24 .40

.29 .71 1.00

4.23 E F G

A 15 12 8 35

B 11 17 19 47

C 21 32 27 80

D 18 13 12 43

65 74 66 205

a) P(GA) = 8/35 = .2286

b) P(BF) = 17/74 = .2297

c) P(CE) = 21/65 = .3231

d) P(EG) = .0000


4.24 C D

A .36 .44 .80

B .11 .09 .20

.47 .53 1.00

a) P(CA) = .36/.80 = .4500

b) P(BD) = .09/.53 = .1698

c) P(AB) = .0000

4.25 Calculator

Yes No

Computer Yes 46 3 49

No 11 15 26

57 18 75

Select a category from each variable and test

P(V1V2) = P(V1).

For example, P(Yes ComputerYes Calculator) = P(Yes Computer)?

75

49

57

46 = ?

.8070 ≠ .6533

Since this is one example that the conditional does not equal the marginal in is matrix, the variable, computer, is not independent of the variable,

calculator.


4.26 Let C = construction Let S = South Atlantic

83,384 total failures10,867 failures in construction

8,010 failures in South Atlantic 1,258 failures in construction and South Atlantic

a) P(S) = 8,010/83,384 = .09606b) P(C ⊥ S) = P(C) + P(S) - P(C _ S) =

10,867/83,384 + 8,010/83,384 - 1,258/83,384 = 17,619/83,384 = .2113

c) P(CS) =

384,83

8010384,83

1258

)(

)( =∩SP

SCP = .15705

d) P(SC) =

384,83

867,10384,83

1258

)(

)( =∩CP

SCP = .11576

e) P(NSNC) = )(

)(1

)(

)(

NCP

SCP

NCP

NCNSP ∪−=∩

but NC = 83,384 - 10,867 = 72,517

and P(NC) = 72,517/83,384 = .869675

Therefore, P(NSNC) = (1 - .2113)/(.869675) = .9069

f) P(NSC) = )(

)()(

)(

)(

CP

SCPCP

CP

CNSP ∩−=∩

but P(C) = 10,867/83,384 = .1303 P(C _ S) = 1,258/83,384 = .0151

Therefore, P(NS C) = (.1303 - .0151)/.1303 = .8842


4.27 Let E = Economy Let Q = QualifiedP(E) = .46 P(Q) = .37 P(E _ Q) = .15a) P(EQ) = P(E _ Q)/P(Q) = .15/.37 = .4054b) P(QE) = P(E _ Q)/P(E) = .15/.46 = .3261c) P(QNE) = P(Q _ NE)/P(NE)

but P(Q _ NE) = P(Q) - P(Q _ E) = .37 - .15 = .22

P(NE) = 1 - P(E) = 1 - .46 = .54

P(QNE) = .22/.54 = .4074d) P(NE _ NQ) = 1 - P(E ⊥ Q) = 1 - [P(E) + P(Q) - P(E _ Q)]

= 1 - [.46 + .37 - .15] = 1 - (.68) = .32

The matrix:

Q

Yes No

E Yes .15 .31 .46

No .22 .32 .54

.37 .63 1.00


4.28 Let EM = email while on phone Let TD = “to-do” lists during meetings

P(EM) = .54 P(TDEM) = .20a) P(EM _ TD) = P(EM) ⋅ P(TDEM) = (.54)(.20) = .1080

b) P(not TDEM) = 1 - P(TDEM) = 1 - .20 = .80c) P(not TD _ EM) = P(EM) - P(EM _ TD) = .54 - .1080 = .4320 Could have been worked as: P(not TD _ EM) = P(EM)⋅ P(not TDEM) = (.54)(.80) = .4320

The matrix:

TD

Yes No

EM

Yes .1080 .4320 .54

No .46

1.00


4.29 Let H = hardware Let S = softwareP(H) = .37 P(S) = .54 P(SH) = .97

a) P(NSH) = 1 - P(SH) = 1 - .97 = .03b) P(SNH) = P(S _ NH)/P(NH)

but P(H _ S) = P(H) ⋅ P(SH) = (.37)(.97) = .3589 so P(NH _ S) = P(S) - P(H _ S) = .54 - .3589 = .1811

P(NH) = 1 - P(H) = 1 - .37 = .63

P(SNH) = (.1811)/(.63) = .2875c) P(NHS) = P(NH _ S)/P(S) = .1811//54 = .3354d) P(NHNS) = P(NH _ NS)/P(NS)

but P(NH _ NS) = P(NH) - P(NH _ S) = .63 - .1811 = .4489

and P(NS) = 1 - P(S) = 1 - .54 = .46

P(NHNS) = .4489/.46 = .9759

The matrix:

S

Yes No

H Yes .3589 .0111 .37

No .1811 .4489 .63

.54 .46 1.00


4.30 Let R = agreed or strongly agreed that lack of role models was a barrier Let S = agreed or strongly agreed that gender-based stereotypes was a barrier

P(R) = .43 P(S) = .46 P(RS) = .77 P(not S) = .54

a.) P(not RS) = 1 - P(RS) = 1 - .77 = .23b.) P(not SR) = P(not S _ R)/P(R)

but P(S _ R) = P(S) ⋅ P(RS) = (.46)(.77) = .3542 so P(not S _ R) = P(R) - P(S _ R) = .43 - .3542 = .0758

Therefore, P(not SR) = (.0758)/(.43) = .1763c.) P(not Rnot S) = P(not R _ not S)/P(not S)

but P(not R _ not S) = P(not S) - P(not S _ R) = .54 - .0758 = .4642

P(not Rnot S) = .4642/.54 = .8596

The matrix:

S

Yes No

R Yes .3542 .0758 .43

No .1058 .4642 .57

.46 .54 1.00


4.31 Let A = product produced on Machine AB = product produces on Machine BC = product produced on Machine CD = defective product

P(A) = .10 P(B) = .40 P(C) = .50 P(DA) = .05 P(DB) = .12 P(DC) = .08

Event Prior Conditional Joint Revised P(Ei)

P(DEi)

P(D _ Ei)

A .10 .05 .005 .005/.093=.0538 B .40 .12 .048 .048/.093=.5161 C .50 .08 .040 .040/.093=.4301

P(D)=.093

Revise: P(AD) = .005/.093 = .0538

P(BD) = .048/.093 = .5161

P(CD) = .040/.093 = .4301


4.32 Let A = Alex fills the orderB = Alicia fills the orderC = Juan fills the orderI = order filled incorrectlyK = order filled correctly

P(A) = .30 P(B) = .45 P(C) = .25P(IA) = .20 P(IB) = .12 P(IC) = .05P(KA) = .80 P(KB) = .88 P(KC) = .95

a) P(B) = .45

b) P(KC) = 1 - P(IC) = 1 - .05 = .95

c)


P(IEi)

P(I _ Ei) P(EiI)

A .30 .20 .0600 .0600/.1265=.4743 B .45 .12 .0540 .0540/.1265=.4269 C .25 .05 .0125 .0125/.1265=.0988

P(I)=.1265

Revised: P(AI) = .0600/.1265 = .4743

P(BI) = .0540/.1265 = .4269

P(CI) = .0125/.1265 = .0988

d)


P(KEi)

P(K _ Ei) P(EiK)

A .30 .80 .2400 .2400/.8735=.2748 B .45 .88 .3960 .3960/.8735=.4533 C .25 .95 .2375 .2375/.8735=.2719

P(K)=.8735


4.33 Let T = lawn treated by Tri-stateG = lawn treated by Green ChemV = very healthy lawnN = not very healthy lawn

P(T) = .72 P(G) = .28 P(VT) = .30 P(VG) = .20


P(VEi)

P(V _ Ei) P(EiV)

A .72 .30 .216 .216/.272=.7941 B .28 .20 .056 .056/.272=.2059

P(V)=.272

Revised: P(TV) = .216/.272 = .7941

P(GV) = .056/.272 = .2059

4.34 Let S = small Let L = large

The prior probabilities are: P(S) = .70 P(L) = .30P(TS) = .18 P(TL) = .82


P(TEi)

P(T _ Ei) P(EiT)

S .70 .18 .1260 .1260/.3720 = .3387 L .30 .82 .2460 .2460/.3720 = .6613

P(T)=.3720

Revised: P(ST) = .1260/.3720 = .3387

P(LT) = .2460/.3720 = .6613

37.2% offer training since P(T) = .3720,.


4.35 Variable 1

D E

A 10 20 30

Variable 2 B 15 5 20

C 30 15 45

55 40 95

a) P(E) = 40/95 = .42105 b) P(B ⊥ D) = P(B) + P(D) - P(B _ D)

= 20/95 + 55/95 - 15/95 = 60/95 = .63158 c) P(A _ E) = 20/95 = .21053

d) P(BE) = 5/40 = .1250 e) P(A ⊥ B) = P(A) + P(B) = 30/95 + 20/95 =

50/95 = .52632 f) P(B _ C) = .0000 (mutually exclusive)

g) P(DC) = 30/45 = .66667

h) P(AB) = P A B

P B

( )

( )

.

/

∩=

0 0 0 0

2 0 9 5 = .0000 (A and B are mutually exclusive)

i) P(A) = P(AD)??

Does 30/95 = 10/95 ??

Since, .31579 ≠ .18182, Variables 1 and 2 are not independent.


4.36

a) P(F _ A) = 7/78 = .08974

b) P(AB) = P A B

P B

( )

( )

.

/

∩=

0 0 0 0

2 2 7 8 = .0000 (A and B are mutually exclusive)

c) P(B) = 22/78 = .28205d) P(E _ F) = .0000 Mutually Exclusive

e) P(DB) = 8/22 = .36364

f) P(BD) = 8/21 = .38095 g) P(D ⊥ C) = 21/78 + 25/78 – 10/78 = 36/78 = .4615

h) P(F) = 16/78 = .20513

D E F G

A 3 9 7 12 31

B 8 4 6 4 22

C 10 5 3 7 25

21 18 16 23 78


4.37

Age(years)

<35 35-44 45-54 55-64 >65

Gender Male .11 .20 .19 .12 .16 .78

Female .07 .08 .04 .02 .01 .22

.18 .28 .23 .14 .17 1.00

a) P(35-44) = .28 b) P(Woman _ 45-54) = .04 c) P(Man ⊥ 35-44) = P(Man) + P(35-44) - P(Man _ 35-44) = .78 + .28 - .20 = .

86 d) P(<35 ⊥ 55-64) = P(<35) + P(55-64) = .18 + .14 = .32 e) P(Woman45-54) = P(Woman _ 45-54)/P(45-54) = .04/.23= .1739

f) P(not W _ not 55-64) = .11 + .20 + .19 + .16 = .66


4.38 Let T = thoroughness Let K = knowledgeP(T) = .78 P(K) = .40 P(T _ K) = .27

a) P(T ⊥ K) = P(T) + P(K) - P(T _ K) =

.78 + .40 - .27 = .91b) P(NT _ NK) = 1 - P(T ⊥ K) = 1 - .91 = .09c) P(KT) = P(K _ T)/P(T) = .27/.78 = .3462d) P(NT _ K) = P(NT) - P(NT _ NK)

but P(NT) = 1 - P(T) = .22 P(NT _ K) = .22 - .09 = .13

The matrix:

K

Yes No

T Yes .27 .51 .78

No .13 .09 .22

.40 .60 1.00


4.39 Let R = retirement Let L = life insurance

P(R) = .42 P(L) = .61 P(R _ L) = .33

a) P(RL) = P(R _ L)/P(L) = .33/.61 = .5410b) P(LR) = P(R _ L)/P(R) = .33/.42 = .7857c) P(L ⊥ R) = P(L) + P(R) - P(L _ R) = .61 + .42 - .33 = .70d) P(R _ NL) = P(R) - P(R _ L) = .42 - .33 = .09e) P(NLR) = P(NL _ R)/P(R) = .09/.42 = .2143

The matrix:

L

Yes No

R Yes .33 .09 .42

No .28 .30 .58

.61 .39 1.00


4.40 P(T) = .16 P(TW) = .20 P(TNE) = .17 P(W) = .21 P(NE) = .20

a) P(W _ T) = P(W)⋅ P(TW) = (.21)(.20) = .042 b) P(NE _ T) = P(NE)⋅ P(TNE) = (.20)(.17) = .034 c) P(WT) = P(W _ T)/P(T) = (.042)/(.16) = .2625d) P(NENT) = P(NE _ NT)/P(NT) = {P(NE)⋅ P(NTNE)}/P(NT)

but P(NTNE) = 1 - P(TNE) = 1 - .17 = .83 and

P(NT) = 1 - P(T) = 1 - .16 = .84

Therefore, P(NENT) = {P(NE)⋅ P(NTNE)}/P(NT) = {(.20)(.83)}/(.84) = .1976

e) P(not W _ not NET) = P(not W _ not NE _ T)/ P(T) but P(not W _ not NE _ T) =

.16 - P(W _ T) - P(NE _ T) = .16 - .042 - .034 = .084 P(not W _ not NE _ T)/ P(T) = (.084)/(.16) = .525


4.41 Let M = MasterCard A = American Express V = Visa

P(M) = .30 P(A) = .20 P(V) = .25P(M _ A) = .08 P(V _ M) = .12 P(A _ V) = .06

a) P(V ⊥ A) = P(V) + P(A) - P(V _ A) = .25 + .20 - .06 = .39 b) P(VM) = P(V _ M)/P(M) = .12/.30 = .40 c) P(MV) = P(V _ M)/P(V) = .12/.25 = .48

d) P(V) = P(VM)??

.25 ≠ .40

Possession of Visa is not independent of possession of MasterCard

e) American Express is not mutually exclusive of Visa because P(A _ V) ≠ .0000


4.42 Let S = believe SS secure N = don't believe SS will be secure <45 = under 45 years old >45 = 45 or more years old

P(N) = .51Therefore, P(S) = 1 - .51 = .49

P(<45) = .57 P(> 45) = .43

P(S>45) = .70

Therefore, P(N>45) = 1 - P(S>45) = 1 - .70 = .30

a) P(>45) = 1 - P(<45) = 1 - .57 = .43

b) P(<45 _ S) = P(S) - P(>45 _ S) = but P(> 45 _ S) = P(> 45)⋅ P(S> 45) = (.43)(.70) = .301 P(<45 _ S) = P(S) - P(>45 _ S) = .49 - .301 = .189

c) P(>45S) = P(>45 _ S)/P(S) = .189/.49 = .6143d) (<45 ⊥ N) = P(<45) + P(N) - P(<45 _ N) =

but P(<45 _ N) = P(<45) - P(<45 _ S) = .57 - .189 = .381so P(<45 ⊥ N) = .57 + .51 - .381 = .699

Probability Matrix Solution for Problem 4.42:

S N

<45 .189 .381 .57

>45 .301 .129 .43

.490 .510 1.00


4.43 Let M = expect to save more R = expect to reduce debt

NM = don't expect to save more NR = don't expect to reduce debt

P(M) = .43 P(R) = .45 P(RM) = .81 P(NRM) = 1 - P(RM) = 1 - .81 = .19 P(NM) = 1 - P(M) = 1 - .43 = .57 P(NR) = 1 - P(R) = 1 - .45 = .55

a) P(M _ R) = P(M)⋅ P(RM) = (.43)(.81) = .3483

b) P(M ⊥ R) = P(M) + P(R) - P(M _ R) = .43 + .45 - .3483 = .5317

c) P(neither save nor reduce debt) = 1 - P(M ⊥ R) = 1 - .5317 = .4683

d) P(M _ NR) = P(M)⋅ P(NRM) = (.43)(.19) = .0817

Probability matrix for problem 4.43:

Reduce

Yes No

Save Yes .3483 .0817 .43

No .1017 .4683 .57

.45 .55 1.00


4.44 Let R = readLet B = checked in the with boss

P(R) = .40 P(B) = .34 P(BR) = .78

a) P(B _ R) = P(R)⋅ P(BR) = (.40)(.78) = .312b) P(NR _ NB) = 1 - P(R ⊥ B)

but P(R ⊥ B) = P(R) + P(B) - P(R _ B) =

.40 + .34 - .312 = .428 P(NR _ NB) = 1 - .428 = .572

c) P(RB) = P(R _ B)/P(B) = (.312)/(.34) = .9176

d) P(NBR) = 1 - P(BR) = 1 - .78 = .22e) P(NBNR) = P(NB _ NR)/P(NR)

but P(NR) = 1 - P(R) = 1 - .40 = .60

P(NBNR) = .572/.60 = .9533

f) Probability matrix for problem 4.44:

B NB

R .312 .088 .40

NR .028 .572 .60

.34 .66 1.00


4.45 Let Q = keep quiet when they see co-worker misconductLet C = call in sick when they are well

P(Q) = .35 P(NQ) = 1 - .35 = .65 P(CQ) = .75 P(QC) = .40a) P(C _ Q) = P(Q)⋅ P(CQ) = (.35)(.75) = .2625b) P(Q ⊥ C) = P(Q) + P(C) - P(C _ Q)

but P(C) must be solved for: P(C _ Q) = P(C) ⋅ P(QC)

.2625 = P(C) ( .40)

Therefore, P(C) = .2625/.40 = .65625 and P(Q ⊥ C) = .35 + .65626 - .2625 = .74375

c) P(NQC) = P(NQ _ C)/P(C) but P(NQ _ C) = P(C) - P(C _ Q) = .65625 - .2625 = ..39375 Therefore, P(NQC) = P(NQ _ C)/P(C) = .39375/.65626 = .60d) P(NQ _ NC) = 1 - P(Q ⊥ C) = 1 - .74375 = .25625e) P(Q _ NC) = P(Q) - P(Q _ C) = .35 - .2625 = .0875


C

Y N

Q Y .2625 .0875 .35

N .39375 .25625 .65

.65625 .34375 1.00


4.46 Let: D = denial I = inappropriate

C = customerP = payment disputeS = specialtyG = delays getting careR = prescription drugs

P(D) = .17 P(I) = .14 P(C) = .14 P(P) = .11P(S) = .10 P(G) = .08 P(R) = .07a) P(P ⊥ S) = P(P) + P(S) = .11 + .10 = .21b) P(R _ C) = .0000 (mutually exclusive)c) P(IS) = P(I _ S)/P(S) = .0000/.10 = .0000d) P(NG _ NP) = 1 - P(G ⊥ P) = 1 - [P(G) + P(P)] =

1 – [.08 + .11] = 1 - .19 = .81


4.47 Let R = retention Let P = process improvementP(R) = .56 P(P _ R) = .36 P(RP) = .90a) P(R _ NP) = P(R) - P(P _ R) = .56 - .36 = .20b) P(PR) = P(P _ R)/P(R) = .36/.56 = .6429

c) P(P) = ?? Solve P(RP) = P(R _ P)/P(P) for P(P): P(P) = P(R _ P)/P(RP) = .36/.90 = .40

d) P(R ⊥ P) = P(R) + P(P) - P(R _ P) =

.56 + .40 - .36 = .60e) P(NR _ NP) = 1 - P(R ⊥ P) = 1 - .60 = .40f) P(RNP) = P(R _ NP)/P(NP)

but P(NP) = 1 - P(P) = 1 - .40 = .60

P(RNP) = .20/.60 = .3333

P

Y N

R Y .36 .20 .56

N .04 .40 .44

.40 .60 1.00

Note: In constructing the matrix, we are given P(R) = .56, P(P _ R) = .36, and

P(RP) = .90. That is, only one marginal probability is given. From P(R), we can get P(NR) by taking 1 - .56 = .44.

However, only these two marginal values can be computed directly. To solve for P(P), using what is given, since we know that 90% of P lies

in the intersection and that the intersection is .36, we can set up an equation to solve for P:

.90P = .36 Solving for P = .40.


4.48 Let M = mail Let S = salesP(M) = .38 P(M _ S) = .0000 P(NM _ NS) = .41

a) P(M _ NS) = P(M) - P(M _ S) = .38 - .00 = .38b) Because P(M _ S) = .0000, P(M ⊥ S) = P(M) + P(S) Therefore, P(S) = P(M ⊥ S) - P(M)

but P(M ⊥ S) = 1 - P(NM _ NS) = 1 - .41 = .59 Thus, P(S) = P(M ⊥ S) - P(M) = .59 - .38 = .21

c) P(SM) = P(S _ M)/P(M) = .0000/.38 = .0000d) P(NMNS) = P(NM _ NS)/P(NS) = .41/.79 = .5190

where: P(NS) = 1 - P(S) = 1 - .21 = .79


S

Y N

M Y .0000 .38 .38

N .21 .41 .62

.21 .79 1.00


4.49 Let F = Flexible Work Let V = Gives time off for VolunteerismP(F) = .41 P(VNF) = .10 P(VF) = .60 from this, P(NF) = 1 - .41 = .59a) P(F ⊥ V) = P(F) + P(V) - P(F _ V) P(F) = .41 and P(F _ V) = P(F)⋅ P(VF) = (.41)(.60) = .246 Find P(V) by using P(V) = P(F _ V) + P(NF _ V) but P(NF _ V) = P(NF)⋅ P(VNF) = (.59)(.10) = .059 so, P(V) = P(F _ V) + P(NF _ V) = .246 + .059 = .305 and P(F ⊥ V) = P(F) + P(V) - P(F _ V) = .41 + .305 - .246 = .469b) P(F _ NV) = P(F) - P(F _ V) = .41 - .246 = .164c) P(FNV) = P(F _ NV)/P(NV) P(F _ NV) = .164

P(NV) = 1 – P(V) = 1 - .305 = .695. P(FNV) = P(F _ NV)/P(NV) = .164/.695 = .2360d) P(NFV) = P(NF _ V)/P(V) = .059/.305 = .1934e) P(NF ⊥ NV) = P(NF) + P(NV) - P(NF _ NV)

P(NF) = .59 P(NV) = .695 Solve for P(NF _ NV) = P(NV) – P(F _ NV) = .695 - .164 = .531 P(NF ⊥ NV) = P(NF) + P(NV) - P(NF _ NV) = .59 + .695 - .531 = .754


V

Y N

F Y .246 .164 .41

N .059 .531 .59

.305 .695 1.00


4.50 Let S = Sarabia Let T = Tran Let J = Jackson Let B = blood test

P(S) = .41 P(T) = .32 P(J) = .27P(BS) = .05 P(BT) = .08 P(BJ) = .06


P(BEi)

P(B _ Ei) P(BiNS)

S .41 .05 .0205 .3291 T .32 .08 .0256 .4109 J .27 .06 .0162 .2600

P(B) = .0623

4.51 Let R = regulations T = tax burden

P(R) = .30 P(T) = .35 P(TR) = .71a) P(R _ T) = P(R)⋅ P(TR) = (.30)(.71) = .2130b) P(R ⊥ T) = P(R) + P(T) - P(R _ T) =

.30 + .35 - .2130 = .4370c) P(R ⊥ T) - P(R _ T) = .4370 - .2130 = .2240d) P(RT) = P(R _ T)/P(T) = .2130/.35 = .6086

e) P(NRT) = 1 - P(RT) = 1 - .6086 = .3914 f) P(NRNT) = P(NR _ NT)/P(NT) = [1 - P(R ⊥ T)]/P(NT) =

(1 - .4370)/.65 = .8662


T

Y N

R Y .213 .087 .30

N .137 .563 .70

.35 .65 1.00


4.52


P(RBEi)

P(RB _ Ei) P(EiRB)

0-24 .353 .11 .03883 .03883/.25066 = .15491

25-34 .142 .24 .03408 .03408/.25066 = .13596

35-44 .160 .27 .04320 .04320/.25066 = .17235

> 45 .345 .39 .13455 .13455/.25066 = .53678

P(RB) = .25066

4.53 Let GH = Good health Let HM = Happy marriage Let FG = Faith in God P(GH) = .29 P(HM) = .21 P(FG) = .40

a) P(HM ⊥ FG) = P(HM) + P(FG) - P(HM _ FG) but P(HM _ FG) = .0000 P(HM ⊥ FG) = P(HM) + P(FG) = .21 + .40 = .61

b) P(HM ⊥ FG ⊥ GH) = P(HM) + P(FG) + P(GH) =

.29 + .21 + .40 = .9000

c) P(FG _ GH) = .0000

The categories are mutually exclusive.

The respondent could not select more than one answer.

d) P(neither FG nor GH nor HM) = 1 - P(HM ⊥ FG ⊥ GH) = 1 - .9000 = .1000

76197039 Ken Black QA 5th Chapter 4 Solution

Documents

Transcript of 76197039 Ken Black QA 5th Chapter 4 Solution