This document provides an overview of the key concepts and objectives covered in Chapter 4 on probability. The chapter aims to help students understand the different ways of assigning probabilities and how to apply probability rules and laws to solve problems. It emphasizes that there are multiple valid approaches to probability problems. The chapter outlines includes topics like classical vs relative frequency vs subjective probabilities, probability rules like addition and multiplication, and conditional probability. It also provides sample problems and their solutions to illustrate the concepts.

1. Chapter 4: Probability 1
Chapter 4
Probability
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 the 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

2. Chapter 4: Probability 2
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 18 presentation of
“revising probabilities in light of sample information (section 18.4).
CHAPTER OUTLINE
4.1 Introduction to Probability
4.2 Methods of Assigning Probabilities
Classical Method of Assigning Probabilities
Relative Frequency of Occurrence
Subjective Probability
4.3 Structure of Probability
Experiment
Event
Elementary Events
Sample Space
Unions and Intersections
Mutually Exclusive Events
Independent Events
Collectively Exhaustive Events
Complimentary Events
Counting 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

3. Chapter 4: Probability 3
4.5 Addition Laws
Probability Matrices
Complement of a Union
Special Law of Addition
4.6 Multiplication Laws
General Law of Multiplication
Special Law of Multiplication
4.7 Conditional Probability
Independent Events
4.8 Revision of Probabilities: Bayes' Rule
KEY TERMS
A Priori Intersection
Bayes' Rule Joint Probability
Classical Method of Assigning Probabilities Marginal Probability
Collectively Exhaustive Events mn Counting Rule
Combinations Mutually Exclusive Events
Complement of a Union Probability Matrix
Complementary Events Relative Frequency of Occurrence
Conditional Probability Sample Space
Elementary Events Set Notation
Event Subjective Probability
Experiment Union
Independent Events Union Probability
SOLUTIONS TO PROBLEMS IN CHAPTER 4
4.1 Enumeration of the six parts: D1, D2, D3, A4, A5, A6
D = Defective part
A = 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

4. Chapter 4: Probability 4
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 part
A = 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.

5. Chapter 4: Probability 5
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 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 = .5000
c) P(D ⊥ E) = P(D) + P(E) = 23/60 + 16/60 = 39/60 = .6500
d) P(C ⊥ F) = P(C) + P(F) - P(C _ F) = 15/60 + 21/60 - 5/60 = 31/60 = .5167

6. Chapter 4: Probability 6
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 - flown in an airplane at least once
T = event - 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.
4.12 P(L) = .75 P(M) = .78 P(M L) = .61
a) 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
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 = .86
b) P(C ⊥ T but not both) = P(C ⊥ T) - P(C _ T) = .86 - .55 = .31

7. Chapter 4: Probability 7
c) P(NC _ NT) = 1 - P(C ⊥ T) = 1 - .86 = .14
d) 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 references
P(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)
Y N
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

8. Chapter 4: Probability 8
.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
a) P(U _ I) = P(U) ⋅ P(I U)
P(U) = .78 P(I) = .15 P(IU) = .11
P(U _ I) = (.78)(.11) = .0858
b) P(U _ NI) = P(U) ⋅ P(NIU) but P(IU) = .11
So, P(NIU) = 1 - .11 = .89 and P(U _ NI) =
P(U) ⋅ P(NIU) = (.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%

9. Chapter 4: Probability 9
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(CS) = .75
a) P(NS) = 1 - .43 = .57
b) P(S _ C) = P(S)⋅ P(CS) = (.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
4.20 Let F = fax machine
Let P = personal computer
Given: P(F) = .10 P(P) = .52 P(PF) = .91
a) P(F _ P) = P(F) ⋅ P(P F) = (.10)(.91) = .091
b) P(F ⊥ P) = P(F) + P(P) - P(F _ P) = .10 + .52 - .091 = .529
c) 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 = .471
e) P(NF _ P) = P(P) - P(F _ P) = .52 - .091 = .429
P NP
F .091 .009 .10
NF .429 .471 .90
.520 .480 1.00
4.21
Let S = safety

10. Chapter 4: Probability 10
Let A = age
P(S) = .30 P(A) = .39 P(A S) = .87
a) P(S _ NA) = P(S)⋅ P(NA S)
but P(NA S) = 1 - P(A S) = 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(A S) = (.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
4.22 Let C = ceiling fans
Let O = outdoor grill
P(C) = .60 P(O) = .29 P(C O) = .13
a) P(C ⊥ O)= P(C) + P(O) - P(C _ O) = .60 + .29 - .13 = .76
b) P(NC _ NO) = 1 - P(C ⊥ O)= 1 - .76 = .24
c) P(NC _ O) = P(O) - P(C _ O) = .29 - .13 = .16
d) P(C _ NO) = P(C) - P(C _ O) = .60 - .13 = .47
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(GA) = 8/35 = .2286
b) P(BF) = 17/74 = .2297

11. Chapter 4: Probability 11
c) P(CE) = 21/65 = .3231
d) P(EG) = .0000
4.24
C D
A .36 .44 .80
B .11 .09 .20
.47 .53 1.00
a) P(CA) = .36/.80 = .4500
b) P(BD) = .09/.53 = .1698
c) P(AB) = .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(V1V2) = P(V1).
For example, P(Yes ComputerYes Calculator) = P(Yes Computer)?
75
49
57
46
= ?
.8070 ≠ .6533
Variable of Computer not independent of Variable of Calculator.

12. Chapter 4: Probability 12
4.26
Let C = construction
Let S = South Atlantic
83,384 total failures
10,867 failures in construction
8,010 failures in South Atlantic
1,258 failures in construction and South Atlantic
a) P(S) = 8,010/83,384 = .09606
b) 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(C  S) =
384,83
8010
384,83
1258
)(
)(
=
∩
SP
SCP
= .15705
d) P(S  C) =
384,83
867,10
384,83
1258
)(
)(
=
∩
CP
SCP
= .11576
e) P(NSNC) =
)(
)(1
)(
)(
NCP
SCP
NCP
NCNSP ∪−
=
∩
but NC = 83,384 - 10,867 = 72,517
and P(NC) = 72,517/83,384 = .869675
Therefore, P(NSNC) = (1 - .2113)/(.869675) = .9069
f) P(NS  C) =
)(
)()(
)(
)(
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 = Qualified

13. Chapter 4: Probability 13
P(E) = .46 P(Q) = .37 P(E _ Q) = .15
a) P(EQ) = P(E _ Q)/P(Q) = .15/.37 = .4054
b) P(QE) = P(E _ Q)/P(E) = .15/.46 = .3261
c) P(QNE) = 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(QNE) = .22/.54 = .4074
d) P(NE _ NQ) = 1 - P(E ⊥ Q) = 1 - [P(E) + P(Q) - P(E _ Q)]
= 1 - [.46 + .37 + .15] = 1 - (.68) = .32
4.28
Let A = airline tickets
Let T = transacting loans
P(A) = .47 P(TA) = .81
a) P(A _ T) = P(A) ⋅ P(TA) = (.47)(.81) = .3807
b) P(NTA) = 1 - P(TA) = 1 - .81 = .19
c) P(NT _ A) = P(A) - P(A _ T) = .47 - .3807 = .0893
4.29 Let H = hardware
Let S = software
P(H) = .37 P(S) = .54 P(SH) = .97
a) P(NSH) = 1 - P(SH) = 1 - .97 = .03
b) P(SNH) = P(S _ NH)/P(NH)
but P(H _ S) = P(H) ⋅ P(SH) = (.37)(.97) = .3589
so P(NH _ S) = P(S) - P(H _ S) = .54 - .3589 = .1811
P(NH) = 1 - P(H) = 1 - .37 = .63
P(SNH) = (.1811)/(.63) = .2875
c) P(NHS) = P(NH _ S)/P(S) = .1811//54 = .3354

14. Chapter 4: Probability 14
d) P(NHNS) = 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(NHNS) = .4489/.46 = .9759
4.30 Let A = product produced on Machine A
B = product produces on Machine B
C = product produced on Machine C
D = defective product
P(A) = .10 P(B) = .40 P(C) = .50
P(DA) = .05 P(DB) = .12 P(DC) = .08
Event Prior Conditional Joint Revised
P(Ei) P(DEi)
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(AD) = .005/.093 = .0538
P(BD) = .048/.093 = .5161
P(CD) = .040/.093 = .4301
4.31 Let A = Alex fills the order
B = Alicia fills the order
C = Juan fills the order
I = order filled incorrectly
K = order filled correctly
P(A) = .30 P(B) = .45 P(C) = .25
P(IA) = .20 P(IB) = .12 P(IC) = .05
P(KA) = .80 P(KB) = .88 P(KC) = .95
a) P(B) = .45
b) P(KC) = 1 - P(IC) = 1 - .05 = .95

15. Chapter 4: Probability 15
c)
Event Prior Conditional Joint Revised
P(Ei) P(IEi)
P(I _ Ei)
P(EiI)
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(AI) = .0600/.1265 = .4743
P(BI) = .0540/.1265 = .4269
P(CI) = .0125/.1265 = .0988
d)
Event Prior Conditional Joint Revised
P(Ei) P(KEi)
P(K _ Ei)
P(EiK)
A .30 .80 .2400 .2400/.8735=.2748
B .45 .88 .3960 .3960/.8735=.4533
C .25 .95 .2375 .2375/.8735=.2719
P(K)=.8735
4.32 Let T = lawn treated by Tri-state
G = lawn treated by Green Chem
V = very healthy lawn
N = not very healthy lawn
P(T) = .72 P(G) = .28 P(VT) = .30 P(VG) = .20
Event Prior Conditional Joint Revised
P(Ei) P(VEi)
P(V _ Ei)
P(EiV)
A .72 .30 .216 .216/.272=.7941
B .28 .20 .056 .056/.272=.2059
P(V)=.272
Revised: P(TV) = .216/.272 = .7941
P(GV) = .056/.272 = .2059

16. Chapter 4: Probability 16
4.33 Let T = training
Let S = small
P(T) = .65
P(ST) = .18 P(NST) = .82
P(SNT) = .75 P(NSNT) = .25
Event Prior Conditional Joint Revised
P(Ei) P(NSEi)
P(NS _ Ei)
P(EiNS)
T .65 .82 .5330 .5330/.6205=.8590
NT .35 .25 .0875 .0875/.6205=.1410
P(NS)=.6205
4.34
Variable 1
D E
A 10 20
Variable 2 B 15 5
C 30 15
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(BE) = 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(DC) = 30/45 = .66667
h) P(AB)= P(A _ B) = .0000 = .0000 mutually exclusive
P(B) 20/95

17. Chapter 4: Probability 17
i) P(A) = P(AD)??
30/95 = 10/95 ??
.31579 ≠ .18182
No, Variables 1 and 2 are not independent.
4.35
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
a) P(F _ A) = 7/78 = .08974
b) P(AB) = P(A _ B) = .0000 = .0000
P(B) 22/78
c) P(B) = 22/78 = .28205
d) P(E _ F) = .0000 Mutually Exclusive
e) P(DB) = 8/22 = .36364
f) P(BD) = 8/21 = .38095
g) P(D ⊥ C) = 21/78 + 25/78 – 10/78 = 36/78 = .4615
h) P(F) = 16/78 = .20513
4.36
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

18. Chapter 4: Probability 18
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(Woman45-54) = P(Woman _ 45-54)/P(45-54) = .04/.23= .1739
f) P(NW _ N 55-64) = .11 + .20 + .19 + .16 = .66
4.37 Let T = thoroughness
Let K = knowledge
P(T) = .78 P(K) = .40 P(T _ K) = .27
a) P(T ⊥ K) = P(T) + P(K) - P(T _ K) =
.78 + .40 - .27 = .91
b) P(NT _ NK) = 1 - P(T ⊥ K) = 1 - .91 = .09
c) P(KT) = P(K _ T)/P(T) = .27/.78 = .3462
d) P(NT _ K) = P(NT) - P(NT _ NK)
but P(NT) = 1 - P(T) = .22
P(NT _ K) = .22 - .09 = .13
4.38 Let R = retirement
Let L = life insurance
P(R) = .42 P(L) = .61 P(R _ L) = .33
a) P(RL) = P(R _ L)/P(L) = .33/.61 = .5410
b) P(LR) = P(R _ L)/P(R) = .33/.42 = .7857
c) P(L ⊥ R) = P(L) + P(R) - P(L _ R) =
.61 + .42 - .33 = .70
d) P(R _ NL) = P(R) - P(R _ L) = .42 - .33 = .09
e) P(NLR) = P(NL _ R)/P(R) = .09/.42 = .2143
4.39 P(T) = .16 P(TW) = .20 P(TNE) = .17
P(W) = .21 P(NE) = .20
a) P(W _ T) = P(W)⋅P(TW) = (.21)(.20) = .042
b) P(NE _ T) = P(NE)⋅P(TNE) = (.20)(.17) = .034
c) P(WT) = P(W _ T)/P(T) = (.042)/(.16) = .2625
d) P(NENT) = P(NE _ NT)/P(NT) = {P(NE)⋅P(NTNE)}/P(NT)
but P(NTNE) = 1 - P(TNE) = 1 - .17 = .83 and

19. Chapter 4: Probability 19
P(NT) = 1 - P(T) = 1 - .16 = .84
Therefore, P(NENT) = {P(NE)⋅P(NTNE)}/P(NT) =
{(.20)(.83)}/(.84) = .1976
e) P(not W _ not NET) = 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
4.40 Let M = Mastercard A = American Express V = Visa
P(M) = .30 P(A) = .20 P(V) = .25
P(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(VM) = P(V _ M)/P(M) = .12/.30 = .40
c) P(MV) = P(V _ M)/P(V) = .12/.25 = .48
d) P(V) = P(VM)??
.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
4.41
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) = .51
Therefore, P(S) = 1 - .51 = .49
P(S>45) = .70
Therefore, P(N>45) = 1 - P(S>45) = 1 - .70 = .30
P(<45) = .57

20. Chapter 4: Probability 20
a) P(>45) = 1 - P(<45) = 1 - .57 = .43
b) P(>45 _ S) = P(>45)⋅P(S>45) = (.57)(.70) = .301
P(<45 _ S) = P(S) - P(>45 _ S) = .49 - .301 = .189
c) P(>45S) = P(>45 _ S)/P(S) = P(>45)⋅P(S>45)/P(S) =
(.43)(.70)/.49 = .6143
d) (<45 ⊥ N) = P(<45) + P(N) - P(<45 _ N) =
but P(<45 _ N) = P(<45) - P(<45 _ S) =
.57 - .189 = .381
so P(<45 ⊥ N) = .57 + .51 - .381 = .699
Probability Matrix Solution for Problem 4.41:
S N
<45 .189 .381 .57
>45 .301 .129 .43
.490 .510 1.00
4.42 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(RM) = .81
P(NRM) = 1-P(RM) = 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(RM) = (.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(NRM) = (.43)(.19) = .0817
Probability matrix for problem 4.42:
Reduce
Yes No

21. Chapter 4: Probability 21
Save
Yes .3483 .0817 .43
No .1017 .4683 .57
.4500 .5500 1.00
4.43 Let R = read
Let B = checked in the with boss
P(R) = .40 P(B) = .34 P(BR) = .78
a) P(B _ R) = P(R)⋅P(BR) = (.40)(.78) = .312
b) 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(RB) = P(R _ B)/P(B) = (.312)/(.34) = .9176
d) P(NBR) = 1 - P(BR) = 1 - .78 = .22
e) P(NBNR) = P(NB _ NR)/P(NR)
but P(NR) = 1 - P(R) = 1 - .40 = .60
P(NBNR) = .572/.60 = .9533
f) Probability matrix for problem 4.43:
B NB
R .312 .088 .40
NR .028 .572 .60
.340 .660 1.00
4.44
Let: D = denial
I = inappropriate
C = customer
P = payment dispute
S = specialty
G = delays getting care

22. Chapter 4: Probability 22
R = prescription drugs
P(D) = .17 P(I) = .14 P(C) = .14 P(P) = .11
P(S) = .10 P(G) = .08 P(R) = .07
a) P(P ⊥ S) = P(P) + P(S) = .11 + .10 = .21
b) P(R _ C) = .0000 (mutually exclusive)
c) P(IS) = P(I _ S)/P(S) = .0000/.10 = .0000
d) P(NG _ NP) = 1 - P(G ⊥ P) = 1 - [P(G) + P(P)] =
1 – [.08 + .11] = 1 - .19 = .81
4.45
Let R = retention
P = process
P(R) = .56 P(P _ R) = .36 P(RP) = .90
a) P(R _ NP) = P(R) - P(P _ R) = .56 - .36 = .20
b) P(PR) = P(P _ R)/P(R) = .36/.56 = .6429
c) P(P) = ??
P(RP) = P(R _ P)/P(P)
so P(P) = P(R _ P)/P(RP) = .36/.90 = .40
d) P(R ⊥ P) = P(R) + P(P) - P(R _ P) =
.56 + .40 - .36 = .60
e) P(NR _ NP) = 1 - P(R ⊥ P) = 1 - .60 = .40
f) P(RNP) = P(R _ NP)/P(NP)
but P(NP) = 1 - P(P) = 1 - .40 = .60
P(RNP) = .20/.60 = .33
4.46 Let M = mail
S = sales
P(M) = .38 P(M _ S) = .0000 P(NM _ NS) = .41
a) P(M _ NS) = P(M) - P(M _ S) = .38 - .00 = .38
b) P(S):
P(M ⊥ S) = 1 - P(NM _ NS) = 1 - .41 = .59
P(M ⊥ S) = P(M) + P(S)
Therefore, P(S) = P(M ⊥ S) - P(M) = .59 - .38 = .21
c) P(SM) = P(S _ M)/P(M) = .00/.38 = .00

23. Chapter 4: Probability 23
d) P(NMNS) = P(NM _ NS)/P(NS) = .41/.79 = .5190
where: P(NS) = 1 - P(S) = 1 - .21 = .79
4.47 Let S = Sarabia
T = Tran
J = Jackson
B = blood test
P(S) = .41 P(T) = .32 P(J) = .27
P(B S) = .05 P(B T) = .08 P(B J) = .06
Event Prior Conditional Joint Revised
P(Ei) P(BEi)
P(B _ Ei)
P(BiNS)
S .41 .05 .0205 .329
T .32 .08 .0256 .411
J .27 .06 .0162 .260
P(B)=.0623
4.48 Let R = regulations
T = tax burden
P(R) = .30 P(T) = .35 P(TR) = .71
a) P(R _ T) = P(R)⋅P(TR) = (.30)(.71) = .2130
b) P(R ⊥ T) = P(R) + P(T) - P(R _ T) =
.30 + .35 - .2130 = .4370
c) P(R ⊥ T) - P(R _ T) = .4370 - .2130 = .2240
d) P(RT) = P(R _ T)/P(T) = .2130/.35 = .6086
e) P(NRT) = 1 - P(RT) = 1 - .6086 = .3914
f) P(NRNT) = P(NR _ NT)/P(NT) = [1 - P(R ⊥ T)]/P(NT) =
(1 - .4370)/.65 = .8662
4.49
Event Prior Conditional Joint Revised
P(Ei) P(NSEi)
P(NS _ Ei)
P(EiNS)
Soup .60 .73 .4380 .8456
Breakfast

24. Chapter 4: Probability 24
Meats .35 .17 .0595 .1149
Hot Dogs .05 .41 .0205 .0396
.5180
4.50 Let GH = Good health
HM = Happy marriage
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