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.
2.
3.
4.
5.
Comprehend the different ways of assigning probability.
Understand and apply marginal, union, joint, and conditional probabilities.
Select the appropriate law of probability to use in solving problems.
Solve problems using the laws of probability, including the law of addition, the law of
multiplication, and the law of conditional probability.
Revise probabilities using Bayes' rule.
CHAPTER OUTLINE
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Introduction to Probability
Methods of Assigning Probabilities
Classical Method of Assigning Probabilities
Relative Frequency of Occurrence
Subjective Probability
Structure of Probability
Experiment
Event
Elementary Events
Sample Space
Unions and Intersections
Mutually Exclusive Events
Independent Events
Collectively Exhaustive Events
Complementary Events
Counting the Possibilities
The mn Counting Rule
Sampling from a Population with Replacement
Combinations: Sampling from a Population without Replacement
Marginal, Union, Joint, and Conditional Probabilities
Addition Laws
Probability Matrices
Complement of a Union
Special Law of Addition
Multiplication Laws
General Law of Multiplication
Special Law of Multiplication
Conditional Probability
Independent Events
Revision of Probabilities: Bayes' Rule
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KEY WORDS
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
STUDY QUESTIONS
1. The _______________ method of assigning probabilities relies on the insight or feelings of the
person determining the probabilities.
2.
If probabilities are determined "a priori" to an experiment using rules and laws, then the
_______________ method of assigning probabilities is being used.
3. The range of possibilities for probability values is from _______________ to _______________.
4.
Suppose a technician keeps track of all defects in raw materials for a single day and uses this
information to determine the probability of finding a defect in raw materials the next day. She is
using the _______________ method of assigning probabilities.
5. The outcome of an experiment is called a(n) _______________. If these outcomes cannot be
decomposed further, then they are referred to as ______________________________.
6.
A computer hardware retailer allows you to order your own computer monitor. The store carries five
different brands of monitors. Each brand comes in 14", 15" or 17" models. In addition, you can
purchase either the deluxe model or the regular model in each brand and in each size. How many
different types of monitors are available considering all the factors? _____________ You probably
used the _______ rule to solve this.
7.
Suppose you are playing the Lotto game and you are trying to “pick” three numbers. For each of the
three numbers, any of the digits 0 through 9 are possible (with replacement). How many different
sets of numbers are available? _____________
8.
A population consists of the odd numbers between 1 and 9 inclusive. If a researcher randomly
samples numbers from the population three at a time, the sample space is _______________. Using
combinations, how could we have determined ahead of time how many elementary events would be
in the sample space? ________
9.
Let A = {2,3,5,6,7,9} and B = {1,3,4,6,7,9}
A  B = _______________ and A ∩ B = _______________.
Chapter 4: Probability
51
10. If the occurrence of one event does not affect the occurrence of the other event, then the events are
said to be _______________.
11. The outcome of the roll of one die is said to be _______________ of the outcome of the roll of
another die.
12. The event of rolling a three on a die and the event of rolling an even number on the same roll with
the same die are _______________.
13.
If the probability of the intersection of two events is zero, then the events are said to be
_______________.
14.
If three objects are selected from a bin, one at a time with replacement, the outcomes of each
selection are _______________.
15.
Suppose a population consists of a manufacturing facility's 1600 workers. Suppose an experiment is
conducted in which a worker is randomly selected. If an event is the selection of a worker over 40
years old, then the event of selecting a worker 40 years or younger is called the _______________
of the first event.
16. The probability of selecting X given that Y has occurred is called a _______________ probability.
17. The probability of X is called a _______________ probability.
18. The probability of X or Y occurring is called a _______________ probability.
19. The probability of X and Y occurring is called a _______________ probability.
20.
Only one of the four types of probability does not use the total possible outcomes in the
denominator when calculating the probability. This type of probability is called _______________
probability.
21.
If the P(AB) = P(A), then the events A, B are _______________ events.
22.
If the P(X) = .53, the P(Y) = .12, and the P(X ∩ Y) = .07, then P(X  Y) = ______________.
23.
If the P(X) = .26, the P(Y) = .31, and X, Y are mutually exclusive, then P(X  Y) =
_______________.
24.
In a company, 47% of the employees wear glasses, 60% of the employees are women, and 28% of
the employees are women and wear glasses. Complete the probability matrix below for this
problem.
Wear Glasses?
Yes
No
Gender
25.
Men
Women
Suppose that in another company, 40% of the workers are part time and 80% of the part time
workers are men. The probability of randomly selecting a company worker who is both part time
and a man is _______________.
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26. The probability of tossing three coins in a row and getting all tails is _______________. This is an
application of the _______________ law of multiplication because each toss is _______________.
27.
Suppose 70% of all cars purchased in America are U.S.A. made and that 18% of all cars purchased
in America are both U.S.A. made and are red. The probability that a randomly selected car
purchased in America is red given that it is U.S.A. made is _______________.
Use the matrix below to answer questions 28-37:
A
B
C
.35
.14
.49
D
.31
.20
.51
.66
.34
1.00
28. The probability of A and C occurring is __________.
29. The probability of A or D occurring is __________.
30. The probability of D occurring is __________.
31. The probability of B occurring given C is __________.
32. The probability of B and D occurring is __________.
33. The probability of C and D occurring is __________.
34. The probability of C or D occurring is __________.
35. The probability of C occurring given D is __________.
36. The probability of C occurring given A is __________.
37. The probability of C or B occurring is __________.
38.
Suppose 42% of all people in a county have characteristic X. Suppose 17% of all people in this
county have characteristic X and characteristic Y. If a person is randomly selected from the county
who is known to have characteristic X, then the probability that they have characteristic Y is
_________________.
39.
Suppose 22% of all parts produced at a plant have flaw X and 37% have flaw Y. In addition,
suppose 53% of the parts with flaw X have flaw Y. If a part is randomly selected, the probability
that it has flaw X or flaw Y is ________________.
40.
Another name for revision of probabilities is _______________.
41.
Suppose the prior probabilities of A and B are .57 and .43 respectively. Suppose that P(EA) = .24
and P(EB) = .56. If E is known to have occurred, then the revised probability of A occurring is
_______________ and of B occurring is _______________.
Chapter 4: Probability
ANSWERS TO STUDY QUESTIONS
1. Subjective
23. .57
2. Classical
24.
3. 0, 1
4. Relative Frequency
Wear Glasses
Yes
No
Men .19
.21 .40
Women .28
.32 .60
.47
.53 1.00
5. Event, Elementary Events
25. .32
6. 30, mn counting rule
26. 1/8 = .125, Special, Independent
7. 103 = 1000 numbers
27. .2571
8. {(1,3,5), (1,3,7), (1,3,9), (1,5,7),
(1,5,9), (1,7,9), (3,5,7), (3,5,9),
(3,7,9), (5,7,9)}, 5C3 = 10
28. .35
9. {1,2,3,4,5,6,7,9}, {3,6,7,9}
30. .51
29. .86
10. Independent
31. .2857
11. Independent
32. .20
12. Mutually Exclusive
33. .0000
13. Mutually Exclusive
34. 1.00
14. Independent
35. .0000
15. Complement
36. .5303
16. Conditional
37. .69
17. Marginal
38. .4048
18. Union
39. .4734
19. Joint or Intersection
40. Bayes’ Rule
20. Conditional
41. .3623, .6377
21. Independent
22. .58
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SOLUTIONS TO ODD-NUMBERED 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,
D1 A6,
D2 D3,
D2 A4,
D2 A5,
D2 A6,
D3 A4,
D3 A5
D3 A6
A4 A5
A4 A6
A5 A6
There are 15 members of the sample space
The probability of selecting exactly one defect out of
two is:
9/15 = .60
4.3
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}
4.5
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.
6C3
=
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.
Chapter 4: Probability
4.7
20C6
=
55
20!
= 38,760
6!14!
It is assumed here that 6 different (without replacement) employees are to be selected.
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
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.
4.13
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
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.
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Solutions Manual and Study Guide
4.15
a)
b)
c)
d)
4.17
P(A ∩
P(D ∩
P(D ∩
P(A ∩
E) =
B) =
E) =
B) =
C
D
E
F
A
5
11
16
8
40
B
2
3
5
7
17
7
14
21
15
57
16/57 =
3/57 =
.0000
.0000
.2807
.0526
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) =
4.19
6 6
36


= .0144
50 50 2500
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
4.21
Let S = safety
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
Chapter 4: Probability
c) P(NS ∩ A) = P(NS) – P(NS ∩ NA)
but P(NS) = 1 – P(S) = 1 – .30 = .70
P(NS ∩ A) = .70 – 571 = .129
4.23
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
c) P(CE) = 21/65 = .3231
d) P(EG) = .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.
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Solutions Manual and Study Guide
4.27
Let E = Economy
Let Q = Qualified
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
4.29
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
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
4.31
Let
P(A) = .30
P(IA) = .20
A = Alex fills the order
B = Alicia fills the order
C = Juan fills the order
I = order filled incorrectly
K = order filled correctly
P(B) = .45 P(C) = .25
P(IB) = .12 P(IC) = .05
Chapter 4: Probability
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
c)
Event
Prior
Conditional
A
B
C
P(Ei)
.30
.45
.25
P(IEi)
.20
.12
.05
Joint
P(I ∩ Ei)
.0600
.0540
.0125
P(I)=.1265
Revised
P(EiI)
.0600/.1265=.4743
.0540/.1265=.4269
.0125/.1265=.0988
Revised: P(AI) = .0600/.1265 = .4743
P(BI) = .0540/.1265 = .4269
P(CI) = .0125/.1265 = .0988
d)
Event
Prior
P(Ei)
A
B
C
.30
.45
.25
4.33
Conditional
P(KEi)
.80
.88
.95
Joint
P(K ∩ Ei)
.2400
.3960
.2375
P(K)=.8735
Revised
P(EiK)
.2400/.8735=.2748
.3960/.8735=.4533
.2375/.8735=.2719
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
P(Ei)
T
NT
.65
.35
Conditional
P(NSEi)
.82
.25
Joint
P(NS ∩ Ei)
.5330
.0875
P(NS)=.6205
Revised
P(EiNS)
.5330/.6205=.8590
.0875/.6205=.1410
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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

22
P( B)
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
4.37
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
4.39
P(T) = .16
P(TW) = .20
P(TNE) = .17
P(W) = .21
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
P(NE) = .20
Chapter 4: Probability
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
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
4.41
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
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:
4.43
S
N
<45
.189
.381
.57
>45
.301
.129
.43
.490
.510
1.00
Let R = read
Let B = checked in the with boss
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Solutions Manual and Study Guide
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
4.45
NB
R
.312
.088
.40
NR
.028
.572
.60
.340
.660
1.00
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
Chapter 4: Probability
4.47 Let S = Sarabia
T = Tran
J = Jackson
B = blood test
P(S) = .41
P(B S) = .05
Event
S
T
J
Prior
P(Ei)
P(T) = .32
P(B T) = .08
P(J) = .27
P(B J) = .06
Conditional
P(BEi)
.41
.32
.27
.05
.08
.06
Joint
P(B ∩ Ei)
Revised
P(BiNS)
.0205
.0256
.0162
P(B)=.0623
.329
.411
.260
4.49
Event
Soup
Breakfast
Meats
Hot Dogs
Prior
Conditional
P(Ei)
.60
P(NSEi)
.73
.35
.05
.17
.41
Joint
P(NS ∩ Ei)
.4380
.0595
.0205
.5180
Revised
P(EiNS)
.8456
.1149
.0396
63
64
Solutions Manual and Study Guide