PROBABILITY AND STATISTICS
FOR ENGINEERS AND
SCIENTISTS 9TH EDITION
WALPOLE SOLUTIONS MANUAL
Contents
1 Introduction to Statistics and Data Analysis 1
2 Probability
11
3 Random Variables and Probability Distributions
27
4 Mathematical Expectation
41
5 Some Discrete Probability Distributions
55
6 Some Continuous Probability Distributions
67
7 Functions of Random Variables
79
8 Fundamental Sampling Distributions and Data Descriptions
85
9 One- and Two-Sample Estimation Problems
97
10 One- and Two-Sample Tests of Hypotheses
113
11 Simple Linear Regression and Correlation
139
12 Multiple Linear Regression and Certain Nonlinear Regression Models
161
13 One-Factor Experiments: General
175
14 Factorial Experiments (Two or More Factors)
197
15 2
k
Factorial Experiments and Fractions
219
16 Nonparametric Statistics
233
17 Statistical Quality Control
247
18 Bayesian Statistics
251
iii
15
Chapter 1
Introduction to Statistics and Data
Analysis
1.1 (a) 15.
(b)
x¯ =
1
(3.4 + 2.5 + 4.8 + ··· + 4.8) = 3.787.
(c) Sample median is the 8th value, after the data is sorted from smallest to largest: 3.6.
(d) A dot plot is shown below.
2.5 3.0 3.5 4.0 4.5 5.0 5.5
(e) After trimming total 40% of the data (20% highest and 20% lowest), the data becomes:
So. the trimmed mean is
2.9 3.0 3.3 3.4 3.6
3.7 4.0 4.4 4.8
1
x¯
tr20
=
9
(2.9 + 3.0 + ··· + 4.8) = 3.678.
(f ) They are about the same.
1.2 (a) Mean=20.7675 and Median=20.610.
(b)
x¯
tr10
= 20.743.
(c) A dot plot is shown below.
18 19 20 21 22 23
(d) No. They are all close to each other.
Copyright ×
c
2012 Pearson Education, Inc. Publishing as Prentice Hall.
1
2 Chapter 1 Introduction to Statistics and Data Analysis 2
Solutions for Exercises in Chapter 1
15−1
1.3 (a) A dot plot is shown below.
200 205 210 215 220 225 230
In the ﬁgure, “×” represents the “No aging” group and “◦ ” represents the “Aging”
group.
(b) Yes; tensile strength is greatly reduced due to the aging process.
(c) Mean
Aging
= 209.90, and Mean
No aging
= 222.10.
(d) Median
Aging
= 210.00, and Median
No aging
= 221.50. The means and medians for each
group are similar to each other.
1.4 (a)
X
¯
A
= 7.950 and X
˜
A
= 8.250;
X
¯
B
= 10.260 and X
˜
B
= 10.150.
(b) A dot plot is shown below.
6.5 7.5 8.5 9.5 10.5 11.5
In the ﬁgure, “×” represents company A and “◦ ” represents company B. The steel rods
made by company B show more ﬂexibility.
1.5 (a) A dot plot is shown below.
−10 0 10 20 30 40
(b)
In the ﬁgure, “×” represents the control group and “◦ ” represents the treatment group.
X
¯
Control
= 5.60, X
˜
Control
= 5.00, and X
¯
tr(10);Control
= 5.13;
X
¯
Treatment
= 7.60, X
˜
Treatment
= 4.50, and X
¯
tr(10);Treatment
= 5.63.
(c) The diﬀerence of the means is 2.0 and the diﬀerences of the medians and the trimmed
means are 0.5, which are much smaller. The possible cause of this might be due to the
extreme values (outliers) in the samples, especially the value of 37.
1.6 (a) A dot plot is shown below.
(b)
1.95 2.05 2.15 2.25 2.35 2.45 2.55
In the ﬁgure, “×” represents the 20
◦
C group and “◦ ” represents the 45
◦
C group.
X
¯
20
◦
C
= 2.1075, and X
¯
45
◦
C
= 2.2350.
(c) Based on the plot, it seems that high temperature yields more high values of tensile
strength, along with a few low values of tensile strength. Overall, the temperature does
have an inﬂuence on the tensile strength.
(d) It also seems that the variation of the tensile strength gets larger when the cure temper-
ature is increased.
1.7 s
2
=
1
[(3.4 − 3.787)
2
+ (2.5 − 3.787)
2
+ (4.8 − 3.787)
2
+ ··· + (4.8 − 3.787)
2
] = 0.94284;
s =
√
s
2
=
√
0.9428 = 0.971.
3 Chapter 1 Introduction to Statistics and Data Analysis 3
Solutions for Exercises in Chapter 1
20−1
1.9 (a) s
10−1
A
B
Control
20
◦
C
45
◦
C
1.8 s
2
=
1
[(18.71 − 20.7675)
2
+ (21.41 − 20.7675)
2
+ ··· + (21.12 − 20.7675)
2
] = 2.5329;
s =
√
2.5345 = 1.5915.
2
No Aging
=
1
[(227 − 222.10)
2
+ (222 − 222.10)
2
+ ··· + (221 − 222.10)
2
] = 23.66;
s
No Aging
=
√
23.62 = 4.86.
s
2 1 2 2 2
Aging
=
10−1
[(219 − 209.90)
s
Aging
=
√
42.12 = 6.49.
+ (214 − 209.90)
+ ··· + (205 − 209.90) ] = 42.10;
(b) Based on the numbers in (a), the variation in “Aging” is smaller that the variation in
“No Aging” although the diﬀerence is not so apparent in the plot.
1.10 For company A: s
2
For company B: s
2
= 1.2078 and s
A
=
√
1.2072 = 1.099.
= 0.3249 and s
B
=
√
0.3249 = 0.570.
1.11 For the control group: s
2
For the treatment group: s
2
= 69.38 and s
Control
= 8.33.
= 128.04 and s
Treatment
= 11.32.
Treatment
1.12 For the cure temperature at 20
◦
C: s
2
For the cure temperature at 45
◦
C: s
2
= 0.005 and s
20
◦
C
= 0.071.
= 0.0413 and s
45
◦
C
= 0.2032.
The variation of the tensile strength is inﬂuenced by the increase of cure temperature.
1.13 (a) Mean = X
¯
= 124.3 and median = X
˜
= 120;
(b) 175 is an extreme observation.
1.14 (a) Mean = X
¯
= 570.5 and median = X
˜
= 571;
(b) Variance = s
2
= 10; standard deviation= s = 3.162; range=10;
(c) Variation of the diameters seems too big so the quality is questionable.
1.15 Yes. The value 0.03125 is actually a P -value and a small value of this quantity means that
the outcome (i.e., HHHHH ) is very unlikely to happen with a fair coin.
1.16 The term on the left side can be manipulated to
n n n
x
i
− nx¯ =
x
i
−
x
i
= 0,
i=1
which is the term on the right side.
i=1 i=1
1.17 (a)
X
¯
smokers
= 43.70 and X
¯
nonsmokers
= 30.32;
(b) s
smokers
= 16.93 and s
nonsmokers
= 7.13;
(c) A dot plot is shown below.
10 20 30 40 50 60 70
4 Chapter 1 Introduction to Statistics and Data Analysis 4
Solutions for Exercises in Chapter 1
In the ﬁgure, “×” represents the nonsmoker group and “◦ ” represents the smoker group.
(d) Smokers appear to take longer time to fall asleep and the time to fall asleep for smoker
group is more variable.
1.18 (a) A stem-and-leaf plot is shown below.
5 Chapter 1 Introduction to Statistics and Data Analysis 5
Solutions for Exercises in Chapter 1
Relative Frequency
Stem Leaf Frequency
1 057 3
2 35 2
3 246 3
4 1138 4
5 22457 5
6 00123445779 11
7 01244456678899 14
8 00011223445589 14
9 0258 4
(b) The following is the relative frequency distribution table.
Relative Frequency Distribution of Grades
Class Interval Class Midpoint Frequency, f Relative Frequency
10 − 19
20 − 29
30 − 39
40 − 49
50 − 59
60 − 69
70 − 79
80 − 89
90 − 99
14.5
24.5
34.5
44.5
54.5
64.5
74.5
84.5
94.5
3
2
3
4
5
11
14
14
4
0.05
0.03
0.05
0.07
0.08
0.18
0.23
0.23
0.07
(c) A histogram plot is given below.
(d)
14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5
Final Exam Grades
The distribution skews to the left.
X
¯
= 65.48, X
˜
= 71.50 and s = 21.13.
1.19 (a) A stem-and-leaf plot is shown below.
Stem Leaf Frequency
0 22233457 8
1 023558 6
2 035 3
3 03 2
4 057 3
6 Chapter 1 Introduction to Statistics and Data Analysis 6
Solutions for Exercises in Chapter 1
5 0569 4
6 0005 4
7 Chapter 1 Introduction to Statistics and Data Analysis 7
Solutions for Exercises in Chapter 1
Relative Frequency
(b) The following is the relative frequency distribution table.
Relative Frequency Distribution of Years
Class Interval Class Midpoint Frequency, f Relative Frequency
0.0 − 0.9
1.0 − 1.9
2.0 − 2.9
3.0 − 3.9
4.0 − 4.9
5.0 − 5.9
6.0 − 6.9
0.45
1.45
2.45
3.45
4.45
5.45
6.45
8
6
3
2
3
4
4
0.267
0.200
0.100
0.067
0.100
0.133
0.133
(c)
X
¯
= 2.797, s = 2.227 and Sample range is 6.5 − 0.2 = 6.3.
1.20 (a) A stem-and-leaf plot is shown next.
Stem Leaf Frequency
0* 34 2
0 56667777777889999 17
1* 0000001223333344 16
1 5566788899 10
2* 034 3
2 7 1
3* 2 1
(b) The relative frequency distribution table is shown next.
Relative Frequency Distribution of Fruit Fly Lives
Class Interval Class Midpoint Frequency, f Relative Frequency
0 − 4
5 − 9
10 − 14
15 − 19
20 − 24
25 − 29
30 − 34
2
7
12
17
22
27
32
2
17
16
10
3
1
1
0.04
0.34
0.32
0.20
0.06
0.02
0.02
(c) A histogram plot is shown next.
2 7 12 17 22 27 32
8 Chapter 1 Introduction to Statistics and Data Analysis 8
Solutions for Exercises in Chapter 1
Fruit fly lives (seconds)
(d)
X
˜
= 10.50.
9 Chapter 1 Introduction to Statistics and Data Analysis 9
Solutions for Exercises in Chapter 1
Relative Frequency
1.21 (a) X
¯
= 74.02 and X
˜
= 78;
(b) s = 39.26.
1.22 (a)
X
¯
= 6.7261 and X
˜
= 0.0536.
(b) A histogram plot is shown next.
6.62 6.66 6.7 6.74 6.78 6.82
Relative Frequency Histogram for Diameter
(c) The data appear to be skewed to the left.
1.23 (a) A dot plot is shown next.
160.15
395.10
(b)
0 100 200 300 400 500 600 700 800 900 1000
X
¯
1980
= 395.1 and X
¯
1990
= 160.2.
(c) The sample mean for 1980 is over twice as large as that of 1990. The variability for
1990 decreased also as seen by looking at the picture in (a). The gap represents an
increase of over 400 ppm. It appears from the data that hydrocarbon emissions decreased
considerably between 1980 and 1990 and that the extreme large emission (over 500 ppm)
were no longer in evidence.
1.24 (a)
X
¯
= 2.8973 and s = 0.5415.
(b) A histogram plot is shown next.
1.8 2.1 2.4 2.7 3 3.3 3.6 3.9
Salaries
7
Solutions for Exercises in Chapter 1
7 Chapter 1 Introduction to Statistics and Data Analysis
wear
Relative Frequency
250 300 350
(c) Use the double-stem-and-leaf plot, we have the following.
Stem Leaf Frequency
1 (84) 1
2* (05)(10)(14)(37)(44)(45) 6
2 (52)(52)(67)(68)(71)(75)(77)(83)(89)(91)(99) 11
3* (10)(13)(14)(22)(36)(37) 6
3 (51)(54)(57)(71)(79)(85) 6
1.25 (a)
(b)
X
¯
= 33.31;
X
˜
= 26.35;
(c) A histogram plot is shown next.
(d)
10 20 30 40 50 60 70 80 90
Percentage of the families
X
¯
tr(10)
= 30.97. This trimmed mean is in the middle of the mean and median using the
full amount of data. Due to the skewness of the data to the right (see plot in (c)), it is
common to use trimmed data to have a more robust result.
1.26 If a model using the function of percent of families to predict staﬀ salaries, it is likely that the
model would be wrong due to several extreme values of the data. Actually if a scatter plot
of these two data sets is made, it is easy to see that some outlier would inﬂuence the trend.
1.27 (a) The averages of the wear are plotted here.
700 800 900 1000 1100 1200 1300
load
8
Solutions for Exercises in Chapter 1
8 Chapter 1 Introduction to Statistics and Data Analysis
(b) When the load value increases, the wear value also increases. It does show certain
relationship.
9
Solutions for Exercises in Chapter 1
9 Chapter 1 Introduction to Statistics and Data Analysis
2.0
2.5
3.0
3.5
wear
100 300 500 700
(c) A plot of wears is shown next.
700 800 900 1000 1100 1200 1300
load
(d) The relationship between load and wear in (c) is not as strong as the case in (a), especially
for the load at 1300. One reason is that there is an extreme value (750) which inﬂuence
the mean value at the load 1300.
1.28 (a) A dot plot is shown next.
High
Low
71.45 71.65 71.85 72.05 72.25 72.45 72.65 72.85 73.05
In the ﬁgure, “×” represents the low-injection-velocity group and “◦ ” represents the
high-injection-velocity group.
(b) It appears that shrinkage values for the low-injection-velocity group is higher than those
for the high-injection-velocity group. Also, the variation of the shrinkage is a little larger
for the low injection velocity than that for the high injection velocity.
1.29 A box plot is shown next.
Solutions for Exercises in Chapter 1
1
0
Chapter 1 Introduction to Statistics and Data Analysis
700 800 900 1000 1100 1200 1300
1.30 A box plot plot is shown next.
1.31 (a) A dot plot is shown next.
Low
High
76 79 82 85 88 91 94
In the ﬁgure, “×” represents the low-injection-velocity group and “◦ ” represents the
high-injection-velocity group.
(b) In this time, the shrinkage values are much higher for the high-injection-velocity group
than those for the low-injection-velocity group. Also, the variation for the former group
is much higher as well.
(c) Since the shrinkage eﬀects change in diﬀerent direction between low mode temperature
and high mold temperature, the apparent interactions between the mold temperature
and injection velocity are signiﬁcant.
1.32 An interaction plot is shown next.
mean shrinkage value
high mold temp
low mold temp
Low
high
injection velocity
1
0
Solutions for Exercises in Chapter 1
1
1
Chapter 1 Introduction to Statistics and Data Analysis
It is quite obvious to ﬁnd the interaction between the two variables. Since in this experimental
data, those two variables can be controlled each at two levels, the interaction can be inves-
1
1
Solutions for Exercises in Chapter 1
1
2
Chapter 1 Introduction to Statistics and Data Analysis
tigated. However, if the data are from an observational studies, in which the variable values
cannot be controlled, it would be diﬃcult to study the interactions among these variables.
1
2
Chapter 2
Probability
2.1 (a) S = {8, 16, 24, 32, 40, 48}.
(b) For x
2
+ 4x − 5 = (x + 5)(x − 1) = 0, the only solutions are x = −5 and x = 1.
S = {−5, 1}.
(c) S = {T, HT, HHT, H HH }.
(d) S = {N. America, S. America, Europe, Asia, Africa, Australia, Antarctica}.
(e) Solving 2x − 4 ≥ 0 gives x ≥ 2. Since we must also have x < 1, it follows that S = φ.
2.2 S = {(x, y) | x
2
+ y
2
< 9; x ≥ 0, y ≥ 0}.
2.3 (a) A = {1, 3}.
(b) B = {1, 2, 3, 4, 5, 6}.
(c) C = {x | x
2
− 4x + 3 = 0} = {x | (x − 1)(x − 3) = 0} = {1, 3}.
(d) D = {0, 1, 2, 3, 4, 5, 6}. Clearly, A = C .
2.4 (a) S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.
(b) S = {(x, y) | 1 ≤ x, y ≤ 6}.
2.5 S = {1HH, 1HT, 1T H, 1T T , 2H, 2T, 3HH, 3HT, 3T H, 3T T , 4H, 4T, 5HH, 5HT, 5T H,
5T T , 6H, 6T }.
2.6 S = {A
1
A
2
, A
1
A
3
, A
1
A
4
, A
2
A
3
, A
2
A
4
, A
3
A
4
}.
2.7 S
1
= {MMMM, MMMF, MMF M, MF MM, F MMM, MMF F, MF MF, M F F M,
FM FM, F FM M, FM M F, M FF F, F M F F, FFM F, FFFM, F FFF }.
S
2
= {0, 1, 2, 3, 4}.
2.8 (a) A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}.
(b) B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2, 1), (2, 3), (2, 4),
(2, 5), (2, 6)}.
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11
12 Chapter 2 Probability 12
Solutions for Exercises in Chapter 2
(c) C = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.
(d) A ∩ C = {(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}.
(e) A ∩ B = φ.
(f ) B ∩ C = {(5, 2), (6, 2)}.
(g) A Venn diagram is shown next.
S
A
A
∩
C
B
B
∩
C
C
2.9 (a) A = {1HH, 1HT, 1T H, 1T T , 2H, 2T }.
(b) B = {1T T , 3T T , 5T T }.
(c) A
= {3HH, 3HT, 3T H, 3T T , 4H, 4T, 5HH, 5HT, 5T H, 5T T , 6H, 6T }.
(d) A
∩ B = {3T T , 5T T }.
(e) A ∪ B = {1HH, 1HT, 1T H, 1T T , 2H, 2T, 3T T , 5T T }.
2.10 (a) S = {F F F, F F N, F NF, N F F, F NN, N F N, NNF, NNN }.
(b) E = {FF F, FFN, F N F, N FF }.
(c) The second river was safe for ﬁshing.
2.11 (a) S = {M
1
M
2
, M
1
F
1
, M
1
F
2
, M
2
M
1
, M
2
F
1
, M
2
F
2
, F
1
M
1
, F
1
M
2
, F
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2012 Pearson Education, Inc. Publishing as Prentice Hall. Copyright ×
c
2012 Pearson Education, Inc. Publishing as Prentice Hall.
1
F
2
, F
2
M
1
, F
2
M
2
,
F
2
F
1
}.
(b) A = {M
1
M
2
, M
1
F
1
, M
1
F
2
, M
2
M
1
, M
2
F
1
, M
2
F
2
}.
(c) B = {M
1
F
1
, M
1
F
2
, M
2
F
1
, M
2
F
2
, F
1
M
1
, F
1
M
2
, F
2
M
1
, F
2
M
2
}.
(d) C = {F
1
F
2
, F
2
F
1
}.
(e) A ∩ B = {M
1
F
1
, M
1
F
2
, M
2
F
1
, M
2
F
2
}.
(f ) A ∪ C = {M
1
M
2
, M
1
F
1
, M
1
F
2
, M
2
M
1
, M
2
F
1
, M
2
F
2
, F
1
F
2
, F
2
F
1
}.
S
A
A ∩B
C
B
13 Chapter 2 Probability 13
Solutions for Exercises in Chapter 2
(g)
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14 Chapter 2 Probability 14
Solutions for Exercises in Chapter 2
2.12 (a) S = {Z Y F, Z N F, W Y F, W N F, SY F, SN F, Z Y M }.
(b) A ∪ B = {Z Y F, Z N F, W Y F, W N F, SY F, SN F } = A.
(c) A ∩ B = {WY F, SY F }.
2.13 A Venn diagram is shown next.
Sample Space
P
S
F
2.14 (a) A ∪ C = {0, 2, 3, 4, 5, 6, 8}.
(b) A ∩ B = φ.
(c) C
= {0, 1, 6, 7, 8, 9}.
(d) C
∩ D = {1, 6, 7}, so (C
∩ D) ∪ B = {1, 3, 5, 6, 7, 9}.
(e) (S ∩ C )
= C
= {0, 1, 6, 7, 8, 9}.
(f ) A ∩ C = {2, 4}, so A ∩ C ∩ D
= {2, 4}.
2.15 (a) A
= {nitrogen, potassium, uranium, oxygen}.
(b) A ∪ C = {copper, sodium, zinc, oxygen}.
(c) A ∩ B
= {copper, zinc} and
C
= {copper, sodium, nitrogen, potassium, uranium, zinc};
so (A ∩ B
) ∪ C
= {copper, sodium, nitrogen, potassium, uranium, zinc}.
(d) B
∩ C
= {copper, uranium, zinc}.
(e) A ∩ B ∩ C = φ.
(f ) A
∪ B
= {copper, nitrogen, potassium, uranium, oxygen, zinc} and
A
∩ C = {oxygen}; so, (A
∪ B
) ∩ (A
∩ C ) = {oxygen}.
2.16 (a) M ∪ N = {x | 0 < x < 9}.
(b) M ∩ N = {x | 1 < x < 5}.
(c) M
∩ N
= {x | 9 < x < 12}.
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15 Chapter 2 Probability 15
Solutions for Exercises in Chapter 2
2.17 A Venn diagram is shown next.
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2012 Pearson Education, Inc. Publishing as Prentice Hall. Copyright ×
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2012 Pearson Education, Inc. Publishing as Prentice Hall.
16 Chapter 2 Probability 16
Solutions for Exercises in Chapter 2
8
S
A B
1 2 3 4
(a) From the above Venn diagram, (A ∩ B)
contains the regions of 1, 2 and 4.
(b) (A ∪ B)
contains region 1.
(c) A Venn diagram is shown next.
S
1
4
B
A
5
2
7
3
C
6
(A ∩ C ) ∪ B contains the regions of 3, 4, 5, 7 and 8.
2.18 (a) Not mutually exclusive.
(b) Mutually exclusive.
(c) Not mutually exclusive.
(d) Mutually exclusive.
2.19 (a) The family will experience mechanical problems but will receive no ticket for traﬃc
violation and will not arrive at a campsite that has no vacancies.
(b) The family will receive a traﬃc ticket and arrive at a campsite that has no vacancies but
will not experience mechanical problems.
(c) The family will experience mechanical problems and will arrive at a campsite that has
no vacancies.
(d) The family will receive a traﬃc ticket but will not arrive at a campsite that has no
vacancies.
(e) The family will not experience mechanical problems.
2.20 (a) 6;
(b) 2;
(c) 2, 5, 6;
(d) 4, 5, 7, 8.
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17 Chapter 2 Probability 17
Solutions for Exercises in Chapter 2
2.21 With n
1
= 6 sightseeing tours each available on n
2
= 3 diﬀerent days, the multiplication rule
gives n
1
n
2
= (6)(3) = 18 ways for a person to arrange a tour.
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18 Chapter 2 Probability 18
Solutions for Exercises in Chapter 2
5
3
2.22 With n
1
= 8 blood types and n
2
= 3 classiﬁcations of blood pressure, the multiplication rule
gives n
1
n
2
= (8)(3) = 24 classiﬁcations.
2.23 Since the die can land in n
1
= 6 ways and a letter can be selected in n
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2
= 26 ways, the
multiplication rule gives n
1
n
2
= (6)(26) = 156 points in S.
2.24 Since a student may be classiﬁed according to n
1
= 4 class standing and n
2
= 2 gender
classiﬁcations, the multiplication rule gives n
1
n
2
= (4)(2) = 8 possible classiﬁcations for the
students.
2.25 With n
1
= 5 diﬀerent shoe styles in n
2
= 4 diﬀerent colors, the multiplication rule gives
n
1
n
2
= (5)(4) = 20 diﬀerent pairs of shoes.
2.26 Using Theorem 2.8, we obtain the followings.
(a) There are
7
= 21 ways.
(b) There are
5
= 10 ways.
2.27 Using the generalized multiplication rule, there are n
1
× n
2
× n
3
× n
4
= (4)(3)(2)(2) = 48
diﬀerent house plans available.
2.28 With n
1
= 5 diﬀerent manufacturers, n
2
= 3 diﬀerent preparations, and n
3
= 2 diﬀerent
strengths, the generalized multiplication rule yields n
1
n
2
n
3
= (5)(3)(2) = 30 diﬀerent ways
to prescribe a drug for asthma.
2.29 With n
1
= 3 race cars, n
2
= 5 brands of gasoline, n
3
= 7 test sites, and n
4
= 2 drivers, the
generalized multiplication rule yields (3)(5)(7)(2) = 210 test runs.
2.30 With n
1
= 2 choices for the ﬁrst question, n
2
= 2 choices for the second question, and so
forth, the generalized multiplication rule yields n
1
n
2
··· n
9
= 2
9
= 512 ways to answer the
test.
2.31 Since the ﬁrst digit is a 5, there are n
1
= 9 possibilities for the second digit and then n
2
= 8
possibilities for the third digit. Therefore, by the multiplication rule there are n
1
n
2
= (9)(8) =
72 registrations to be checked.
2.32 (a) By Theorem 2.3, there are 6! = 720 ways.
(b) A certain 3 persons can follow each other in a line of 6 people in a speciﬁed order is 4
ways or in (4)(3!) = 24 ways with regard to order. The other 3 persons can then be
placed in line in 3! = 6 ways. By Theorem 2.1, there are total (24)(6) = 144 ways to
line up 6 people with a certain 3 following each other.
(c) Similar as in (b), the number of ways that a speciﬁed 2 persons can follow each other in
a line of 6 people is (5)(2!)(4!) = 240 ways. Therefore, there are 720 − 240 = 480 ways
if a certain 2 persons refuse to follow each other.
19 Chapter 2 Probability 19
Solutions for Exercises in Chapter 2
2.33 (a) With n
1
= 4 possible answers for the ﬁrst question, n
2
= 4 possible answers for the
second question, and so forth, the generalized multiplication rule yields 4
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5
= 1024 ways
to answer the test.
20 Chapter 2 Probability 20
Solutions for Exercises in Chapter 2
5!
3!
2!
(b) With n
1
= 3 wrong answers for the ﬁrst question, n
2
= 3 wrong answers for the second
question, and so forth, the generalized multiplication rule yields
n
1
n
2
n
3
n
4
n
5
= (3)(3)(3)(3)(3) = 3
5
= 243
ways to answer the test and get all questions wrong.
2.34 (a) By Theorem 2.3, 7! = 5040.
(b) Since the ﬁrst letter must be m, the remaining 6 letters can be arranged in 6! = 720
ways.
2.35 The ﬁrst house can be placed on any of the n
1
= 9 lots, the second house on any of the
remaining n
2
= 8 lots, and so forth. Therefore, there are 9! = 362, 880 ways to place the 9
homes on the 9 lots.
2.36 (a) Any of the 6 nonzero digits can be chosen for the hundreds position, and of the remaining
6 digits for the tens position, leaving 5 digits for the units position. So, there are
(6)(6)(5) = 180 three digit numbers.
(b) The units position can be ﬁlled using any of the 3 odd digits. Any of the remaining 5
nonzero digits can be chosen for the hundreds position, leaving a choice of 5 digits for
the tens position. By Theorem 2.2, there are (3)(5)(5) = 75 three digit odd numbers.
(c) If a 4, 5, or 6 is used in the hundreds position there remain 6 and 5 choices, respectively,
for the tens and units positions. This gives (3)(6)(5) = 90 three digit numbers beginning
with a 4, 5, or 6. If a 3 is used in the hundreds position, then a 4, 5, or 6 must be
used in the tens position leaving 5 choices for the units position. In this case, there are
(1)(3)(5) = 15 three digit number begin with a 3. So, the total number of three digit
numbers that are greater than 330 is 90 + 15 = 105.
2.37 The ﬁrst seat must be ﬁlled by any of 5 girls and the second seat by any of 4 boys. Continuing
in this manner, the total number of ways to seat the 5 girls and 4 boys is (5)(4)(4)(3)(3)(2)(2)(1)(1) =
2880.
2.38 (a) 8! = 40320.
(b) There are 4! ways to seat 4 couples and then each member of a couple can be interchanged
resulting in 2
4
(4!) = 384 ways.
(c) By Theorem 2.3, the members of each gender can be seated in 4! ways. Then using
Theorem 2.1, both men and women can be seated in (4!)(4!) = 576 ways.
2.39 (a) Any of the n
1
= 8 ﬁnalists may come in ﬁrst, and of the n
2
= 7 remaining ﬁnalists can
then come in second, and so forth. By Theorem 2.3, there 8! = 40320 possible orders in
which 8 ﬁnalists may ﬁnish the spelling bee.
(b) The possible orders for the ﬁrst three positions are
8
P
3
=
8!
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= 336.
2.40 By Theorem 2.4,
8
P
5
=
8!
= 6720.
2.41 By Theorem 2.4,
6
P
4
=
6!
= 360.
21 Chapter 2 Probability 21
Solutions for Exercises in Chapter 2
37!
2.42 By Theorem 2.4,
40
P
3
=
40!
= 59, 280.
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22 Chapter 2 Probability 22
Solutions for Exercises in Chapter 2
3!2!
3
3!4!2!
3
18
36
2.43 By Theorem 2.5, there are 4! = 24 ways.
2.44 By Theorem 2.5, there are 7! = 5040 arrangements.
2.45 By Theorem 2.6, there are
8!
= 3360.
2.46 By Theorem 2.6, there are
9!
= 1260 ways.
2.47 By Theorem 2.8, there are
8
= 56 ways.
2.48 Assume February 29th as March 1st for the leap year. There are total 365 days in a year.
The number of ways that all these 60 students will have diﬀerent birth dates (i.e, arranging
60 from 365) is
365
P
60
. This is a very large number.
2.49 (a) Sum of the probabilities exceeds 1.
(b) Sum of the probabilities is less than 1.
(c) A negative probability.
(d) Probability of both a heart and a black card is zero.
2.50 Assuming equal weights
(a) P (A) =
5
;
(b) P (C ) =
1
;
(c) P (A ∩ C ) =
7
.
2.51 S = {$10, $25, $100} with weights 275/500 = 11/20, 150/500 = 3/10, and 75/500 = 3/20,
respectively. The probability that the ﬁrst envelope purchased contains less than $100 is
equal to 11/20 + 3/10 = 17/20.
2.52 (a) P (S ∩ D
) = 88/500 = 22/125.
(b) P (E ∩ D ∩ S
) = 31/500.
(c) P (S
∩ E
) = 171/500.
2.53 Consider the events
S: industry will locate in Shanghai,
B: industry will locate in Beijing.
(a) P (S ∩ B) = P (S) + P (B) − P (S ∪ B) = 0.7 + 0.4 − 0.8 = 0.3.
(b) P (S
∩ B
) = 1 − P (S ∪ B) = 1 − 0.8 = 0.2.
2.54 Consider the events
B: customer invests in tax-free bonds,
M : customer invests in mutual funds.
(a) P (B ∪ M ) = P (B) + P (M ) − P (B ∩ M ) = 0.6 + 0.3 − 0.15 = 0.75.
(b) P (B
∩ M
) = 1 − P (B ∪ M ) = 1 − 0.75 = 0.25.
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23 Chapter 2 Probability 23
Solutions for Exercises in Chapter 2
N
= .
2.55 By Theorem 2.2, there are N = (26)(25)(24)(9)(8)(7)(6) = 47, 174, 400 possible ways to code
the items of which n = (5)(25)(24)(8)(7)(6)(4) = 4, 032, 000 begin with a vowel and end with
an even digit. Therefore,
n
10
117
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24 Chapter 2 Probability 24
Solutions for Exercises in Chapter 2
5
)
5
)
(
9
(
9
3
)
2.56 (a) Let A = Defect in brake system; B = Defect in fuel system; P (A ∪ B) = P (A) + P (B) −
P (A ∩ B) = 0.25 + 0.17 − 0.15 = 0.27.
(b) P (No defect) = 1 − P (A ∪ B) = 1 − 0.27 = 0.73.
2.57 (a) Since 5 of the 26 letters are vowels, we get a probability of 5/26.
(b) Since 9 of the 26 letters precede j, we get a probability of 9/26.
(c) Since 19 of the 26 letters follow g, we get a probability of 19/26.
2.58 (a) Of the (6)(6) = 36 elements in the sample space, only 5 elements (2,6), (3,5), (4,4), (5,3),
and (6,2) add to 8. Hence the probability of obtaining a total of 8 is then 5/36.
(b) Ten of the 36 elements total at most 5. Hence the probability of obtaining a total of at
most is 10/36=5/18.
4 48
2.59 (a)
(
3
)(
2
)
=
94
.
(
52
54145
13 13
(b)
(
4
)(
1
)
=
143
(
52
39984
.
1 8
2.60 (a)
(
1
)(
2
)
=
1
.
3
)
5 3
(b)
(
2
)(
1
)
=
5
14
.
3
2.61 (a) P (M ∪ H ) = 88/100 = 22/25;
(b) P (M
∩ H
) = 12/100 = 3/25;
(c) P (H ∩ M
) = 34/100 = 17/50.
2.62 (a) 9;
(b) 1/9.
2.63 (a) 0.32;
(b) 0.68;
(c) oﬃce or den.
2.64 (a) 1 − 0.42 = 0.58;
(b) 1 − 0.04 = 0.96.
2.65 P (A) = 0.2 and P (B) = 0.35
(a) P (A
) = 1 − 0.2 = 0.8;
(b) P (A
∩ B
) = 1 − P (A ∪ B) = 1 − 0.2 − 0.35 = 0.45;
(c) P (A ∪ B) = 0.2 + 0.35 = 0.55.
2.66 (a) 0.02 + 0.30 = 0.32 = 32%;
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20 Chapter 2 Probability 20
Solutions for Exercises in Chapter 2
(b) 0.32 + 0.25 + 0.30 = 0.87 = 87%;
(c) 0.05 + 0.06 + 0.02 = 0.13 = 13%;
(d) 1 − 0.05 − 0.32 = 0.63 = 63%.
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19
Solutions for Exercises in Chapter 2
19 Chapter 2 Probability
2.67 (a) 0.12 + 0.19 = 0.31;
(b) 1 − 0.07 = 0.93;
(c) 0.12 + 0.19 = 0.31.
2.68 (a) 1 − 0.40 = 0.60.
(b) The probability that all six purchasing the electric oven or all six purchasing the gas oven
is 0.007 + 0.104 = 0.111. So the probability that at least one of each type is purchased
is 1 − 0.111 = 0.889.
2.69 (a) P (C ) = 1 − P (A) − P (B) = 1 − 0.990 − 0.001 = 0.009;
(b) P (B
) = 1 − P (B) = 1 − 0.001 = 0.999;
(c) P (B) + P (C ) = 0.01.
2.70 (a) ($4.50 − $4.00) × 50, 000 = $25, 000;
(b) Since the probability of underﬁlling is 0.001, we would expect 50, 000 × 0.001 = 50 boxes
to be underﬁlled. So, instead of having ($4.50 − $4.00) × 50 = $25 proﬁt for those 50
boxes, there are a loss of $4.00 × 50 = $200 due to the cost. So, the loss in proﬁt expected
due to underﬁlling is $25 + $200 = $250.
2.71 (a) 1 − 0.95 − 0.002 = 0.048;
(b) ($25.00 − $20.00) × 10, 000 = $50, 000;
(c) (0.05)(10, 000) × $5.00 + (0.05)(10, 000) × $20 = $12, 500.
2.72 P (A
∩ B
) = 1 − P (A ∪ B) = 1 − (P (A) + P (B) − P (A ∩ B) = 1 + P (A ∩ B) − P (A) − P (B).
2.73 (a) The probability that a convict who pushed dope, also committed armed robbery.
(b) The probability that a convict who committed armed robbery, did not push dope.
(c) The probability that a convict who did not push dope also did not commit armed robbery.
2.74 P (S | A) = 10/18 = 5/9.
2.75 Consider the events:
M : a person is a male;
S: a person has a secondary education;
C : a person has a college degree.
(a) P (M | S) = 28/78 = 14/39;
(b) P (C
| M
) = 95/112.
2.76 Consider the events:
A: a person is experiencing hypertension,
B: a person is a heavy smoker,
C : a person is a nonsmoker.
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20
Solutions for Exercises in Chapter 2
20 Chapter 2 Probability
(a) P (A | B) = 30/49;
(b) P (C | A
) = 48/93 = 16/31.
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21
Solutions for Exercises in Chapter 2
21 Chapter 2 Probability
(d)
0.102+0.046
2.77 (a) P (M ∩ P ∩ H ) =
10
=
5
;
68 34
(b) P (H ∩ M | P
) =
P (H ∩M ∩P )
=
22−10
=
12
=
3
.
P (P
) 100−68 32 8
2.78 (a) (0.90)(0.08) = 0.072;
(b) (0.90)(0.92)(0.12) = 0.099.
2.79 (a) 0.018;
(b) 0.22 + 0.002 + 0.160 + 0.102 + 0.046 + 0.084 = 0.614;
(c) 0.102/0.614 = 0.166;
0.175+0.134
= 0.479.
2.80 Consider the events:
C : an oil change is needed,
F : an oil ﬁlter is needed.
(a) P (F | C ) =
P (F ∩C )
=
0.14
= 0.56.
P (C ) 0.25
(b) P (C | F ) =
P (C ∩F )
=
0.14
= 0.35.
P (F )
2.81 Consider the events:
0.40
H : husband watches a certain show,
W : wife watches the same show.
(a) P (W ∩ H ) = P (W )P (H | W ) = (0.5)(0.7) = 0.35.
(b) P (W | H ) =
P (W ∩H )
=
0.35
= 0.875.
P (H )
0.4
(c) P (W ∪ H ) = P (W ) + P (H ) − P (W ∩ H ) = 0.5 + 0.4 − 0.35 = 0.55.
2.82 Consider the events:
H : the husband will vote on the bond referendum,
W : the wife will vote on the bond referendum.
Then P (H ) = 0.21, P (W ) = 0.28, and P (H ∩ W ) = 0.15.
(a) P (H ∪ W ) = P (H ) + P (W ) − P (H ∩ W ) = 0.21 + 0.28 − 0.15 = 0.34.
(b) P (W | H ) =
P (H ∩W )
=
0.15
=
5
.
P (H )
0.21 7
(c) P (H | W
) =
P (H ∩W )
=
0.06
=
1
.
2.83 Consider the events:
P (W
) 0.72 12
A: the vehicle is a camper,
B: the vehicle has Canadian license plates.
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22
Solutions for Exercises in Chapter 2
22 Chapter 2 Probability
(a) P (B | A) =
P (A∩B)
=
0.09
=
9
.
P (A) 0.28 28
(b) P (A | B) =
P (A∩B)
=
0.09
=
3
.
P (B) 0.12 4
(c) P (B
∪ A
) = 1 − P (A ∩ B) = 1 − 0.09 = 0.91.
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23
Solutions for Exercises in Chapter 2
23 Chapter 2 Probability
(
8
8
2.84 Deﬁne
H : head of household is home,
C : a change is made in long distance carriers.
P (H ∩ C ) = P (H )P (C | H ) = (0.4)(0.3) = 0.12.
2.85 Consider the events:
A: the doctor makes a correct diagnosis,
B: the patient sues.
P (A
∩ B) = P (A
)P (B | A
) = (0.3)(0.9) = 0.27.
2.86 (a) 0.43;
(b) (0.53)(0.22) = 0.12;
(c) 1 − (0.47)(0.22) = 0.90.
2.87 Consider the events:
A: the house is open,
B: the correct key is selected.
1 7
P (A) = 0.4, P (A
) = 0.6, and P (B) =
(
1
)(
2
)
=
3
= 0.375.
3
)
So, P [A ∪ (A
∩ B)] = P (A) + P (A
)P (B) = 0.4 + (0.6)(0.375) = 0.625.
2.88 Consider the events:
F : failed the test,
P : passed the test.
(a) P (failed at least one tests) = 1 − P (P
1
P
2
P
3
P
4
) = 1 − (0.99)(0.97)(0.98)(0.99) = 1 −
0.93 = 0.07,
(b) P (failed 2 or 3) = 1 − P (P
2
P
3
) = 1 − (0.97)(0.98) = 0.0494.
(c) 100 × 0.07 = 7.
(d) 0.25.
2.89 Let A and B represent the availability of each ﬁre engine.
(a) P (A
∩ B
) = P (A
)P (B
) = (0.04)(0.04) = 0.0016.
(b) P (A ∪ B) = 1 − P (A
∩ B
) = 1 − 0.0016 = 0.9984.
2.90 (a) P (A ∩ B ∩ C ) = P (C | A ∩ B)P (B | A)P (A) = (0.20)(0.75)(0.3) = 0.045.
(b) P (B
∩ C ) = P (A ∩ B
∩ C ) + P (A
∩ B
∩ C ) = P (C | A ∩ B
)P (B
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| A)P (A) + P (C | A
∩
B
)P (B
| A
)P (A
) = (0.80)(1 − 0.75)(0.3) + (0.90)(1 − 0.20)(1 − 0.3) = 0.564.
(c) Use similar argument as in (a) and (b), P (C ) = P (A ∩ B ∩ C ) + P (A ∩ B
∩ C ) + P (A
∩
B ∩ C ) + P (A
∩ B
∩ C ) = 0.045 + 0.060 + 0.021 + 0.504 = 0.630.
(d) P (A | B
∩ C ) = P (A ∩ B
∩ C )/P (B
∩ C ) = (0.06)(0.564) = 0.1064.
24
Solutions for Exercises in Chapter 2
24 Chapter 2 Probability
2.91 (a) P (Q
1
∩ Q
2
∩ Q
3
∩ Q
4
) = P (Q
1
)P (Q
2
| Q
1
)P (Q
3
| Q
1
∩ Q
2
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)P (Q
4
| Q
1
∩ Q
2
∩ Q
3
) =
(15/20)(14/19)(13/18)(12/17) = 91/323.
25
Solutions for Exercises in Chapter 2
25 Chapter 2 Probability
91
P system works
=
= 0.2045.
P (B
1
| A) =
(0.005)(0.20)
(b) Let A be the event that 4 good quarts of milk are selected. Then
15
P (A) =
4
= .
20
4
2.92 P = (0.95)[1 − (1 − 0.7)(1 − 0.8)](0.9) = 0.8037.
2.93 This is a parallel system of two series subsystems.
323
(a) P = 1 − [1 − (0.7)(0.7)][1 − (0.8)(0.8)(0.8)] = 0.75112.
(b) P =
P (A
∩C ∩D∩E) (0.3)(0.8)(0.8)(0.8)
0.75112
2.94 Deﬁne S: the system works.
P (A
| S
) =
P (A ∩S )
=
P (A )(1−P (C
∩D∩E))
(0.3)[1−(0.8)(0.8)(0.8)]
P (S
) 1−P (S)
=
1−0.75112
= 0.588.
2.95 Consider the events:
C : an adult selected has cancer,
D: the adult is diagnosed as having cancer.
P (C ) = 0.05, P (D | C ) = 0.78, P (C
) = 0.95 and P (D | C
) = 0.06. So, P (D) = P (C ∩ D) +
P (C
∩ D) = (0.05)(0.78) + (0.95)(0.06) = 0.096.
2.96 Let S
1
, S
2
, S
3
, and S
4
represent the events that a person is speeding as he passes through
the respective locations and let R represent the event that the radar traps is operating
resulting in a speeding ticket. Then the probability that he receives a speeding ticket:
4
P (R) =
P (R | S
i
)P (S
i
) = (0.4)(0.2) + (0.3)(0.1) + (0.2)(0.5) + (0.3)(0.2) = 0.27.
i=1
2.97 P (C | D) =
P (C ∩D)
=
0.039
= 0.40625.
P (D) 0.096
2.98 P (S
2
| R) =
P (R∩ S
2
)
=
0.03
= 1/9.
P (R)
2.99 Consider the events:
A: no expiration date,
0.27
B
1
: John is the inspector, P (B
1
) = 0.20 and P (A | B
1
) = 0.005,
B
2
: Tom is the inspector, P (B
2
) = 0.60 and P (A | B
2
) = 0.010,
B
3
: Jeﬀ is the inspector, P (B
3
) = 0.15 and P (A | B
3
) = 0.011,
B
4
: Pat is the inspector, P (B
4
) = 0.05 and P (A | B
4
) = 0.005,
(0.005)(0.20)+(0.010)(0.60)+(0.011)(0.15)+(0.005)(0.05)
= 0.1124.
2.100 Consider the events
E: a malfunction by other human errors,
A: station A, B: station B, and C : station C .
P (C | E) =
P (E | C )P (C )
(5/10)(10/43)
0.1163
P (E | A)P (A)+P (E | B)P (B)+P (E | C )P (C )
=
(7/18)(18/43)+(7/15)(15/43)+(5/10)(10/43)
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=
26
Solutions for Exercises in Chapter 2
26 Chapter 2 Probability
0.4419
= 0.2632.
2.101 Consider the events:
A: a customer purchases latex paint,
A
: a customer purchases semigloss paint,
B: a customer purchases rollers.
P (A | B) =
P (B | A)P (A)
(0.60)(0.75)
P (B | A)P (A)+P (B | A
)P (A
)
=
(0.60)(0.75)+(0.25)(0.30)
= 0.857.
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2012 Pearson Education, Inc. Publishing as Prentice Hall.
27
Solutions for Exercises in Chapter 2
27 Chapter 2 Probability
1
2.102 If we use the assumptions that the host would not open the door you picked nor the door
with the prize behind it, we can use Bayes rule to solve the problem. Denote by events A, B,
and C , that the prize is behind doors A, B, and C , respectively. Of course P (A) = P (B) =
P (C ) = 1/3. Denote by H the event that you picked door A and the host opened door B,
while there is no prize behind the door B. Then
P (A|H ) =
P (H |B)P (B)
P (H |A)P (A) + P (H |B)P (B) + P (H |C )P (C )
=
P (H |B)
=
P (H |A) + P (H |B) + P (H |C )
1/2 1
= .
0 + 1/2+1 3
Hence you should switch door.
2.103 Consider the events:
G: guilty of committing a crime,
I : innocent of the crime,
i: judged innocent of the crime,
g: judged guilty of the crime.
P (I | g) =
P (g | I )P (I )
(0.01)(0.95)
P (g | G)P (G)+P (g | I )P (I )
=
(0.05)(0.90)+(0.01)(0.95)
= 0.1743.
2.104 Let A
i
be the event that the ith patient is allergic to some type of week.
(a) P (A
1
∩ A
2
∩ A
3
∩ A
4
) + P (A
1
∩ A
2
∩ A
3
∩ A
4
) + P (A
1
∩ A
2
∩ A
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3
∩ A
4
) + P (A
1
∩ A
2
∩ A
3
∩
A
4
) = P (A
1
)P (A
2
)P (A
3
)P (A
)+P (A
1
)P (A
2
)P (A
)P (A
4
)+P (A
1
)P (A
)P (A
3
)P (A
4
)+
4 3 2
P (A
)P (A
2
)P (A
3
)P (A
4
) = (4)(1/2)
4
= 1/4.
(b) P (A
∩ A
∩ A
∩ A
) = P (A
)P (A
)P (A
)P (A
) = (1/2)
4
= 1/16.
1 2 3 4 1 2 3 4
2.105 No solution necessary.
2.106 (a) 0.28 + 0.10 + 0.17 = 0.55.
(b) 1 − 0.17 = 0.83.
(c) 0.10 + 0.17 = 0.27.
2.107 The number of hands =
13
13
13
13
.
4 6 1 2
2.108 (a) P (M
1
∩ M
2
∩ M
3
∩ M
4
) = (0.1)
4
= 0.0001, where M
i
represents that ith person make a
mistake.
(b) P (J ∩ C ∩ R
∩ W
) = (0.1)(0.1)(0.9)(0.9) = 0.0081.
2.109 Let R, S, and L represent the events that a client is assigned a room at the Ramada Inn,
Sheraton, and Lakeview Motor Lodge, respectively, and let F represents the event that the
plumbing is faulty.
28
Solutions for Exercises in Chapter 2
28 Chapter 2 Probability
(a) P (F ) = P (F | R)P (R) + P (F | S)P (S) + P (F | L)P (L) = (0.05)(0.2) + (0.04)(0.4) +
(0.08)(0.3) = 0.054.
(b) P (L | F ) =
(0.08)(0.3)
4
0.054
=
9
.
2.110 Denote by R the event that a patient survives. Then P (R) = 0.8.
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29
Solutions for Exercises in Chapter 2
29 Chapter 2 Probability
1
0.0680
= 0.2941.
2
)
2
)
(a) P (R
1
∩R
2
∩R
3
)+P (R
1
∩R
2
∩R
3
)P (R
1
∩R
2
∩R
3
) = P (R
1
)P (R
2
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)P (R
3
)+P (R
1
)P (R
2
)P (R
3
)+
P (R
)P (R
2
)P (R
3
) = (3)(0.8)(0.8)(0.2) = 0.384.
(b) P (R
1
∩ R
2
∩ R
3
) = P (R
1
)P (R
2
)P (R
3
) = (0.8)
3
= 0.512.
2.111 Consider events
M : an inmate is a male,
N : an inmate is under 25 years of age.
P (M
∩ N
) = P (M
) + P (N
) − P (M
∪ N
) = 2/5 + 1/3 − 5/8 = 13/120.
2.112 There are
4
5
6
= 800 possible selections.
3 3 3
2.113 Consider the events:
B
i
: a black ball is drawn on the ith drawl,
G
i
: a green ball is drawn on the ith drawl.
(a) P (B
1
∩ B
2
∩ B
3
) + P (G
1
∩ G
2
∩ G
3
) = (6/10)(6/10)(6/10) + (4/10)(4/10)(4/10) = 7/25.
(b) The probability that each color is represented is 1 − 7/25 = 18/25.
2.114 The total number of ways to receive 2 or 3 defective sets among 5 that are purchased is
3
9
+
3
9
= 288.
2 3 3 2
2.115 Consider the events:
O: overrun,
A: consulting ﬁrm A,
B: consulting ﬁrm B,
C : consulting ﬁrm C .
(a) P (C | O) =
P (O | C )P (C )
(0.15)(0.25)
0.0375
P (O | A)P (A)+P (O | B)P (B)+P (O | C )P (C )
=
(0.05)(0.40)+(0.03)(0.35)+(0.15)(0.25)
=
0.0680
= 0.5515.
(b) P (A | O) =
(0.05)(0.40)
2.116 (a) 36;
(b) 12;
(c) order is not important.
2.117 (a)
1
= 0.0016;
(
36
12 24
(b)
(
1
)(
1
)
=
288
(
36
630
= 0.4571.
2.118 Consider the events:
C : a woman over 60 has the cancer,
P : the test gives a positive result.
So, P (C ) = 0.07, P (P
| C ) = 0.1 and P (P | C
) = 0.05.
P (C | P
) =
P (P | C )P (C )
(0.1)(0.07)
0.007
30
Solutions for Exercises in Chapter 2
30 Chapter 2 Probability
P (P
| C )P (C )+P (P
| C )P (C )
=
(0.1)(0.07)+(1−0.05)(1−0.07)
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=
0.8905
= 0.00786.
2.119 Consider the events:
A: two nondefective components are selected,
N : a lot does not contain defective components, P (N ) = 0.6, P (A | N ) = 1,
31
Solutions for Exercises in Chapter 2
31 Chapter 2 Probability
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19
2
)
2
)
0.9505
= 0.2841;
3
P (A)
=
= 0.25.
O: a lot contains one defective component, P (O) = 0.3, P (A | O) =
(
2
)
=
9
,
(
20
10
18
T : a lot contains two defective components,P (T ) = 0.1, P (A | T ) =
(
2
)
=
153
.
(
20
190
(a) P (N | A) =
P (A | N )P (N )
(1)(0.6)
=
0.6
P (A | N )P (N )+P (A | O)P (O)+P (A | T )P (T )
=
(1)(0.6)+(9/10)(0.3)+(153/190)(0.1)
0.9505
= 0.6312;
(b) P (O | A) =
(9/10)(0.3)
(c) P (T | A) = 1 − 0.6312 − 0.2841 = 0.0847.
2.120 Consider events:
D: a person has the rare disease, P (D) = 1/500,
P : the test shows a positive result, P (P | D) = 0.95 and P (P | D
) = 0.01.
P (D | P ) =
P (P | D)P (D)
(0.95)(1/500)
P (P | D)P (D)+P (P | D
)P (D
)
=
(0.95)(1/500)+(0.01)(1−1/500)
= 0.1599.
2.121 Consider the events:
1: engineer 1, P (1) = 0.7, and 2: engineer 2, P (2) = 0.3,
E: an error has occurred in estimating cost, P (E | 1) = 0.02 and P (E | 2) = 0.04.
P (1 | E) =
P (E | 1)P (1)
(0.02)(0.7)
P (E | 1)P (1)+P (E | 2)P (2)
=
(0.02)(0.7)+(0.04)(0.3)
= 0.5385, and
P (2 | E) = 1 − 0.5385 = 0.4615. So, more likely engineer 1 did the job.
2.122 Consider the events: D: an item is defective
(a) P (D
1
D
2
D
3
) = P (D
1
)P (D
2
)P (D
3
) = (0.2)
3
= 0.008.
(b) P (three out of four are defectives) =
4
(0.2)
3
(1 − 0.2) = 0.0256.
2.123 Let A be the event that an injured worker is admitted to the hospital and N be the event
that an injured worker is back to work the next day. P (A) = 0.10, P (N ) = 0.15 and
P (A ∩ N ) = 0.02. So, P (A ∪ N ) = P (A) + P (N ) − P (A ∩ N ) = 0.1 + 0.15 − 0.02 = 0.23.
2.124 Consider the events:
T : an operator is trained, P (T ) = 0.5,
M an operator meets quota, P (M | T ) = 0.9 and P (M | T
) = 0.65.
P (T | M ) =
P (M | T )P (T )
(0.9)(0.5)
P (M | T )P (T )+P (M | T
)P (T
)
=
(0.9)(0.5)+(0.65)(0.5)
= 0.581.
2.125 Consider the events:
A: purchased from vendor A,
D: a customer is dissatisﬁed.
Then P (A) = 0.2, P (A | D) = 0.5, and P (D) = 0.1.
So, P (D | A) =
P (A | D)P (D)
(0.5)(0.1)
0.2
2.126 (a) P (Union member | New company (same ﬁeld)) =
13
=
13
= 0.5652.
13+10 23
(b) P (Unemployed | Union member) =
2
=
2
= 0.034.
32
Solutions for Exercises in Chapter 2
32 Chapter 2 Probability
2.127 Consider the events:
C : the queen is a carrier, P (C ) = 0.5,
40+13+4+2 59
D: a prince has the disease, P (D | C ) = 0.5.
P (D
D
D
| C )P (C )
(0.5)
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3
(0.5)
1
P (C | D
1
D
2
D
3
) =
1 2 3
=
(0.5)
3
(0.5)+1(0.5)
=
9
.
P (D
D
D
| C )P (C )+P (D
D
D
| C
)P (C
)
1 2 3 1 2 3
Probability and Statistics for Engineers and Scientists 9th Edition Walpole Solutions Manual