Solucionario Walpole 8 ED
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Solucionario Walpole 8 ED

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system works.
P (A\u2032 | S \u2032) = P (A\u2032\u2229S\u2032)
P (S\u2032)
= P (A
\u2032)(1\u2212P (C\u2229D\u2229E))
1\u2212P (S) =
(0.3)[1\u2212(0.8)(0.8)(0.8)]
1\u22120.75112 = 0.588.
2.101 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 \u2032) = 0.95 and P (D | C \u2032) = 0.06. So, P (D) =
P (C \u2229D) + P (C \u2032 \u2229D) = (0.05)(0.78) + (0.95)(0.06) = 0.096.
24 Chapter 2 Probability
2.102 Let S1, S2, S3, and S4 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:
P (R) =
4\u2211
i=1
P (R | Si)P (Si) = (0.4)(0.2) + (0.3)(0.1) + (0.2)(0.5) + (0.3)(0.2) = 0.27.
2.103 P (C | D) = P (C\u2229D)
P (D)
= 0.039
0.096
= 0.40625.
2.104 P (S2 | R) = P (R\u2229 S2)P (R) = 0.030.27 = 1/9.
2.105 Consider the events:
A: no expiration date,
B1: John is the inspector, P (B1) = 0.20 and P (A | B1) = 0.005,
B2: Tom is the inspector, P (B2) = 0.60 and P (A | B2) = 0.010,
B3: Jeff is the inspector, P (B3) = 0.15 and P (A | B3) = 0.011,
B4: Pat is the inspector, P (B4) = 0.05 and P (A | B4) = 0.005,
P (B1 | A) = (0.005)(0.20)(0.005)(0.20)+(0.010)(0.60)+(0.011)(0.15)+(0.005)(0.05) = 0.1124.
2.106 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)
P (E | A)P (A)+P (E | B)P (B)+P (E | C)P (C) =
(5/10)(10/43)
(7/18)(18/43)+(7/15)(15/43)+(5/10)(10/43)
=
0.1163
0.4419
= 0.2632.
2.107 (a) P (A \u2229 B \u2229 C) = P (C | A \u2229B)P (B | A)P (A) = (0.20)(0.75)(0.3) = 0.045.
(b) P (B\u2032 \u2229 C) = P (A \u2229B\u2032 \u2229 C) + P (A\u2032 \u2229B\u2032 \u2229 C) = P (C | A \u2229B\u2032)P (B\u2032 | A)P (A) +
P (C | A\u2032\u2229B\u2032)P (B\u2032 | A\u2032)P (A\u2032) = (0.80)(1\u22120.75)(0.3)+(0.90)(1\u22120.20)(1\u22120.3) =
0.564.
(c) Use similar argument as in (a) and (b), P (C) = P (A\u2229B \u2229C)+P (A\u2229B\u2032 \u2229C)+
P (A\u2032 \u2229 B \u2229 C) + P (A\u2032 \u2229B\u2032 \u2229 C) = 0.045 + 0.060 + 0.021 + 0.504 = 0.630.
(d) P (A | B\u2032 \u2229 C) = P (A \u2229B\u2032 \u2229 C)/P (B\u2032 \u2229 C) = (0.06)(0.564) = 0.1064.
2.108 Consider the events:
A: a customer purchases latex paint,
A\u2032: a customer purchases semigloss paint,
B: a customer purchases rollers.
P (A | B) = P (B | A)P (A)
P (B | A)P (A)+P (B | A\u2032)P (A\u2032) =
(0.60)(0.75)
(0.60)(0.75)+(0.25)(0.30)
= 0.857.
2.109 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)
P (g | G)P (G)+P (g | I)P (I) =
(0.01)(0.95)
(0.05)(0.90)+(0.01)(0.95)
= 0.1743.
Solutions for Exercises in Chapter 2 25
2.110 Let Ai be the event that the ith patient is allergic to some type of week.
(a) P (A1 \u2229 A2 \u2229 A3 \u2229 A\u20324) + P (A1 \u2229 A2 \u2229 A\u20323 \u2229 A4) + P (A1 \u2229 A\u20322 \u2229 A3 \u2229 A4) +
P (A
\u2032
1 \u2229 A2 \u2229 A3 \u2229 A4) = P (A1)P (A2)P (A3)P (A\u20324) + P (A1)P (A2)P (A\u20323)P (A4) +
P (A1)P (A
\u2032
2)P (A3)P (A4) + P (A
\u2032
1)P (A2)P (A3)P (A4) = (4)(1/2)
4 = 1/4.
(b) P (A
\u2032
1 \u2229A\u20322 \u2229A\u20323 \u2229 A\u20324) = P (A\u20321)P (A\u20322)P (A\u20323)P (A\u20324) = (1/2)4 = 1/16.
2.111 No solution necessary.
2.112 (a) 0.28 + 0.10 + 0.17 = 0.55.
(b) 1\u2212 0.17 = 0.83.
(c) 0.10 + 0.17 = 0.27.
2.113 P =
(134 )(
13
6 )(
13
1 )(
13
2 )
(5213)
.
2.114 (a) P (M1 \u2229M2 \u2229M3 \u2229M4) = (0.1)4 = 0.0001, where Mi represents that ith person
make a mistake.
(b) P (J \u2229 C \u2229R\u2032 \u2229W \u2032) = (0.1)(0.1)(0.9)(0.9) = 0.0081.
2.115 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.
(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)
0.054
= 4
9
.
2.116 (a) There are
(
9
3
)
= 84 possible committees.
(b) There are
(
4
1
)(
5
2
)
= 40 possible committees.
(c) There are
(
3
1
)(
1
1
)(
5
1
)
= 15 possible committees.
2.117 Denote by R the event that a patient survives. Then P (R) = 0.8.
(a) P (R1 \u2229 R2 \u2229 R\u20323) + P (R1 \u2229 R\u20322 \u2229 R3)P (R\u20321 \u2229 R2 \u2229 R3) = P (R1)P (R2)P (R\u20323) +
P (R1)P (R
\u2032
2)P (R3) + P (R
\u2032
1)P (R2)P (R3) = (3)(0.8)(0.8)(0.2) = 0.384.
(b) P (R1 \u2229R2 \u2229 R3) = P (R1)P (R2)P (R3) = (0.8)3 = 0.512.
2.118 Consider events
M : an inmate is a male,
N : an inmate is under 25 years of age.
P (M \u2032 \u2229N \u2032) = P (M \u2032) + P (N \u2032)\u2212 P (M \u2032 \u222aN \u2032) = 2/5 + 1/3\u2212 5/8 = 13/120.
2.119 There are
(
4
3
)(
5
3
)(
6
3
)
= 800 possible selections.
26 Chapter 2 Probability
2.120 Consider the events:
Bi: a black ball is drawn on the ith drawl,
Gi: a green ball is drawn on the ith drawl.
(a) P (B1\u2229B2\u2229B3)+P (G1\u2229G2\u2229G3) = (6/10)(6/10)(6/10)+(4/10)(4/10)(4/10) =
7/25.
(b) The probability that each color is represented is 1\u2212 7/25 = 18/25.
2.121 The total number of ways to receive 2 or 3 defective sets among 5 that are purchased
is
(
3
2
)(
9
3
)
+
(
3
3
)(
9
2
)
= 288.
2.122 A Venn diagram is shown next.
A
B
C
S
1
2
3
4
5
6
7
8
(a) (A \u2229B)\u2032: 1, 2, 3, 6, 7, 8.
(b) (A \u222aB)\u2032: 1, 6.
(c) (A \u2229 C) \u222a B: 3, 4, 5, 7, 8.
2.123 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)
P (O | A)P (A)+P (O | B)P (B)+P (O | C)P (C) =
(0.15)(0.25)
(0.05)(0.40)+(0.03)(0.35)+(0.15)(0.25)
=
0.0375
0.0680
= 0.5515.
(b) P (A | O) = (0.05)(0.40)
0.0680
= 0.2941.
2.124 (a) 36;
(b) 12;
(c) order is not important.
2.125 (a) 1
(362 )
= 0.0016;
(b)
(121 )(
24
1 )
(362 )
= 288
630
= 0.4571.
Solutions for Exercises in Chapter 2 27
2.126 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 \u2032 | C) = 0.1 and P (P | C \u2032) = 0.05.
P (C | P \u2032) = P (P \u2032 | C)P (C)
P (P \u2032 | C)P (C)+P (P \u2032 | C\u2032)P (C\u2032) =
(0.1)(0.07)
(0.1)(0.07)+(1\u22120.05)(1\u22120.07) =
0.007
0.8905
= 0.00786.
2.127 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,
O: a lot contains one defective component, P (O) = 0.3, P (A | O) = (
19
2 )
(202 )
= 9
10
,
T : a lot contains two defective components,P (T ) = 0.1, P (A | T ) = (
18
2 )
(202 )
= 153
190
.
(a) P (N | A) = P (A | N)P (N)
P (A | N)P (N)+P (A | O)P (O)+P (A | T )P (T ) =
(1)(0.6)
(1)(0.6)+(9/10)(0.3)+(153/190)(0.1)
= 0.6
0.9505
= 0.6312;
(b) P (O | A) = (9/10)(0.3)
0.9505
= 0.2841;
(c) P (T | A) = 1\u2212 0.6312\u2212 0.2841 = 0.0847.
2.128 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\u2032) = 0.01.
P (D | P ) = P (P | D)P (D)
P (P | D)P (D)+P (P | D\u2032)P (D\u2032) =
(0.95)(1/500)
(0.95)(1/500)+(0.01)(1\u22121/500) = 0.1599.
2.129 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)
P (E | 1)P (1)+P (E | 2)P (2) =
(0.02)(0.7)
(0.02)(0.7)+(0.04)(0.3)
= 0.5385, and
P (2 | E) = 1\u2212 0.5385 = 0.4615. So, more likely engineer 1 did the job.
2.130 Consider the events: D: an item is defective
(a) P (D1D2D3) = P (D1)P (D2)P (D3) = (0.2)
3 = 0.008.
(b) P (three out of four are defectives) =
(
4
3
)
(0.2)3(1\u2212 0.2) = 0.0256.
2.131 LetA be the event that an injured worker is admitted to the hospital andN 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\u2229N) = 0.02. So, P (A\u222aN) = P (A)+P (N)\u2212P (A\u2229N) = 0.1+0.15\u22120.02 = 0.23.
2.132 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 \u2032) = 0.65.
P (T | M) = P (M | T )P (T )
P