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
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
Chapter 4: Probability
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
3
KEY TERMS
A Priori
Bayes' Rule
Classical Method of Assigning Probabilities
Collectively Exhaustive Events
Combinations
Complement of a Union
Complementary Events
Conditional Probability
Elementary Events
Event
Experiment
Independent Events
Intersection
Joint Probability
Marginal Probability
mn Counting Rule
Mutually Exclusive Events
Probability Matrix
Relative Frequency of Occurrence
Sample Space
Set Notation
Subjective Probability
Union
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,
D1 D3,
D1 A4,
D1 A5,
D2 D3,
D2 A4,
D2 A5,
D2 A6,
D3 A5
D3 A6
A4 A5
A4 A6
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 A4,
D1 A1 A4,
D1 A3 A4,
D2 A1 A4,
D2 A3 A4,
A1 A3 A4,
D1 D2 A2,
D1 A1 A2,
D1 A2 A3,
D2 A1 A2,
D2 A2 A3,
A1 A2 A3,
A2 A3 A4
D1 D2 A3,
D1 A1 A3,
D1 A2 A4,
D2 A1 A3,
D2 A2 A4,
A1 A2 A4,
Combinations are used to counting the sample space because sampling is done
without replacement.
Chapter 4: Probability
C3 =
6
5
6!
= 20
3!3!
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
20
C6 =
20!
= 38,760
6!14!
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
Chapter 4: Probability
6
4.10
a)
b)
c)
d)
4.11
E
F
A
.10
.03
.13
B
.04
.12
.16
C
.27
.06
.33
D
.31
.07
.38
.72
.28
1.00
P(A  F) = P(A) + P(F) - P(A  F) = .13 + .28 - .03 = .38
P(E  B) = P(E) + P(B) - P(E  B) = .72 + .16 - .04 = .84
P(B  C) = P(B) + P(C) =.16 + .33 =
.49
P(E  F) = P(E) + P(F) = .72 + .28 =
1.000
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
4.13
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
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
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 =
b) P(F  T) - P(F  T) = .63 - .35 = .28
c) 1 - P(F  T) = 1 - .63 = .37
d)
Y
Y
.35
N
.19
.54
N
.09
.37
.46
.44
.56
1.00
4.15
a)
b)
c)
d)
P(A 
P(D 
P(D 
P(A 
C
D
E
F
A
5
11
16
8
40
B
2
3
5
7
17
7
14
21
15
57
E) =
B) =
E) =
B) =
16/57 =
3/57 =
.0000
.0000
.2807
.0526
4.16
D
E
F
A
.12
.13
.08
.33
B
.18
.09
.04
.31
C
.06
.24
.06
.36
.63
Chapter 4: Probability
.36
.46
.18
8
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) =
6 5
30


= .0122
50 49 2450
b) (with replacement)
P(D1  D2) = P(D1)  P(D2) =
4.18
6 6
36


= .0144
50 50 2500
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
I
No
Yes
No
.15
.85
.78
.22
The answer to a) is found in the YES-YES cell. To compute this cell, take 11%
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
4.21
Let S = safety
P
NP
F
.091
.009
.10
NF
.429
.471
.90
.520
.480
1.00
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
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
Computer
Yes
No
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)?
46 49

?
57 75
.8070  .6533
Variable of Computer not independent of Variable of Calculator.
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
1258
P(C  S ) 83,384
c) P(C  S) =
= .15705

8010
P( S )
83,384
1258
P(C  S ) 83,384
d) P(S  C) =
= .11576

10,867
P(C )
83,384
e) P(NSNC) =
P( NS  NC ) 1  P(C  S )

P( NC )
P( NC )
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) =
P( NS  C ) P(C )  P(C  S )

P(C )
P(C )
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
Chapter 4: Probability
P(E) = .46
P(Q) = .37
13
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
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
Event
A
B
C
Prior
Conditional
P(Ei)
.10
.40
.50
P(DEi)
.05
.12
.08
Joint
P(D  Ei)
.005
.048
.040
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
P(DC) = .08
Revised
.005/.093=.0538
.048/.093=.5161
.040/.093=.4301
Chapter 4: Probability
15
c)
Event
Prior
Conditional
A
B
C
P(Ei)
.30
.45
.25
P(IEi)
.20
.12
.05
Joint
P(I  Ei)
Revised
P(EiI)
.0600/.1265=.4743
.0540/.1265=.4269
.0125/.1265=.0988
.0600
.0540
.0125
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
A
B
C
P(Ei)
.30
.45
.25
P(KEi)
.80
.88
.95
4.32
Let
P(T) = .72
Joint
P(K  Ei)
Revised
.2400
.3960
.2375
P(K)=.8735
P(EiK)
.2400/.8735=.2748
.3960/.8735=.4533
.2375/.8735=.2719
T = lawn treated by Tri-state
G = lawn treated by Green Chem
V = very healthy lawn
N = not very healthy lawn
P(G) = .28 P(VT) = .30
Event
Prior
Conditional
A
B
P(Ei)
.72
.28
P(VEi)
.30
.20
Joint
P(V  Ei)
.216
.056
P(V)=.272
Revised: P(TV) = .216/.272 = .7941
P(GV) = .056/.272 = .2059
P(VG) = .20
Revised
P(EiV)
.216/.272=.7941
.056/.272=.2059
Chapter 4: Probability
4.33
16
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
T
NT
P(Ei)
.65
.35
P(NSEi)
.82
.25
Joint
P(NS  Ei)
Revised
P(EiNS)
.5330/.6205=.8590
.0875/.6205=.1410
.5330
.0875
P(NS)=.6205
4.34
Variable 1
Variable 2
D
E
A
10
20
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
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)
Gender
<35
35-44
45-54
55-64
>65
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
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(W) = .21 P(NE) = .20
P(TNE) = .17
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
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(M  A) = .08
P(V) = .25
P(V  M) = .12
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
P(A  V) = .06
Chapter 4: Probability
20
a)
b)
P(>45) = 1 - P(<45) = 1 - .57 = .43
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:
4.42
S
N
<45
.189
.381
.57
>45
.301
.129
.43
.490
.510
1.00
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
Chapter 4: Probability
21
Yes
.3483
.0817
.43
No
.1017
.4683
.57
.4500
.5500
1.00
Save
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
Chapter 4: Probability
22
R = prescription drugs
P(D) = .17
P(S) = .10
P(I) = .14
P(G) = .08
P(C) = .14
P(R) = .07
P(P) = .11
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
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(B S) = .05
Event
S
T
J
4.48
P(T) = .32
P(J) = .27
P(B T) = .08 P(B J) = .06
Prior
Conditional
P(Ei)
.41
.32
.27
P(BEi)
.05
.08
.06
Joint
P(B  Ei)
.0205
.0256
.0162
P(B)=.0623
Revised
P(BiNS)
.329
.411
.260
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
Soup
Breakfast
Prior
Conditional
P(Ei)
.60
P(NSEi)
.73
Joint
P(NS  Ei)
.4380
Revised
P(EiNS)
.8456
Chapter 4: Probability
Meats
Hot Dogs
4.50
.35
.05
.17
.41
24
.0595
.0205
.5180
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
.1149
.0396