Chapter 2 Homework Part 2 Answers
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Industrial Engineering
Date
Dec 6, 2023
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Chapter 2 Homework Part 2
ANSWERS
1. (Section 2.5)
Samples of skin experiencing desquamation are analyzed for both moisture and melanin
content. The results from 100 skin samples are as follows:
Let A denote the event that a sample has low melanin content, and let B denote the event that a sample has high
moisture content. Determine the following probabilities:
a.
𝑃(?)
b.
𝑃(?)
c.
𝑃(?|?)
d.
𝑃(?|?)
ANSWERS:
(?) 𝑃(?) =
7 + 32
100
= 0.39 (?) 𝑃(?) =
13 + 7
100
= 0.20
(?) 𝑃(?|?) =
7/10
20/100
= 0.35 (?) 𝑃(?) =
7/100
39/100
= 0.1795
2. (Section 2.5)
A maintenance firm has gathered the following information regarding the failure mechanisms
for air conditioning systems:
The units without evidence of gas leaks or electrical failure showed other types of failure. If this is a
representative sample of AC failure, find the probability
a.
that failure involves a gas leak.
b.
that there is evidence of electrical failure given that there was a gas leak.
c.
that there is evidence of a gas leak given that there is evidence of electrical failure.
ANSWERS:
(a) 𝑃(gas leak) =
55 + 32
107
= 0.813 (b) 𝑃(electric failure|gas leak) =
55/107
87/102
= 0.632
(c) 𝑃(gas leak|electric failure) =
55/107
72/107
= 0.764
3. (Section 2.6)
In the 2012 presidential election, exit polls from the critical state of Ohio provided the
following results.
What is the probability a randomly selected respondent voted for Obama?
ANSWER:
𝑃(?) = 𝑃(?)𝑃(?|?) + 𝑃(?
′
)𝑃(?|?
′
) = 0.40(0.47) + 0.60(0.52) = 0.50
4. (Section 2.6)
A lot of 100 semiconductor chips contain 20 that are defective.
a.
Two are selected, at random, without replacement, from the lot. Determine the probability that the second
chip selected is defective.
b.
Three are selected, at random, without replacement, from the lot. Determine the probability that all are
defective.
ANSWERS:
Let
?
and
?
denote the events that the first and second chips selected are defective, respectively.
(?) 𝑃(?) = 𝑃(?|?)𝑃(?) + 𝑃(?|?
′
)𝑃(?′) =
19
99
(
20
100
) +
20
99
(
80
100
) = 0.2
Let C denote the event that the third chip selected is defective.
(?) 𝑃(? ∩ ? ∩ ?) = 𝑃(?|? ∩ ?)𝑃(? ∩ ?) = 𝑃(?|? ∩ ?)𝑃(?|?)𝑃(?) =
18
98
(
19
99
) (
20
100
) = 0.00705
5. (Section 2.7)
Samples of emissions from three suppliers are classified for conformance to air-quality
specifications. The results from 100 samples are summarized as follows:
Let A denote the event that a sample is from supplier 1, and let B deonote the event that a sample conforms to
specifications.
a.
Are events A and B independent.
b.
Determine
𝑃(?|?),
ANSWERS:
(a) The events
?
and
?
are not independent.
𝑃(? ∩ ?) ≠ 𝑃(?)𝑃(?): 𝑃(? ∩ ?) =
22
100
,
𝑃(?) =
30
100
, 𝑃(?) =
77
100
(?) 𝑃(?|?) =
𝑃(? ∩ ?)
𝑃(?)
=
22/100
30/100
= 0.733
6. (Section 2.7)
The following circuit operates if and only if there is a path of functional devices from left to
right. The probability that each device functions is shown below.
Assume that the probability that a device is functional does not depend on whether or not other devices are
functional. What is the probabilty that the circuit operates?
𝑃(?) = (0.9)(0.8)(0.7) = 0.504 𝑃(?) = (0.95)(0.95)(0.95) = 0.8574
𝑃(? ∩ ?) = 𝑃(?)𝑃(?) = (0.504)(0.8574) = 0.4321
The probability the circuit operates:
𝑃(? ∪ ?) = 𝑃(?) + 𝑃(?) − 𝑃(? ∩ ?) = 0.504 + 0.8574 − 0.4321 = 0.9293
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7. (Section 2.8)
Software to detect fraud in consumer phone cards tracks the number of metropolitan areas
where calls originate each day. It is found that 1% of the legitimate users originate calls from two or more
metropolitan areas in a single day. However, 30% of fraudulent users originate calls from two or more
metropolitan areas in a single day. The proportion of fraudulent users is 0.01%. If the same user originates calls
from two or more metropolitan areas in a single day, what is the probability that the user is fraudulent?
Let F denote a fraudulent user and let T denote a user that originates calls from two or more metropolitan areas
in a day
ANSWER:
003
.
0
)
9999
(.
01
.
0
)
0001
.
0
(
30
.
0
)
0001
.
0
(
30
.
0
)
'
(
)
'
(
)
(
)
(
)
(
)
(
)
(
=
+
=
+
=
F
P
F
T
P
F
P
F
T
P
F
P
F
T
P
T
F
P
A Tree diagram might help with the inventory:
8. (Section 2.8)
A recreational equipment supplier finds that among orders that include tents, 40% also include
sleeping mats. Only 5% of orders that do not include tents do include sleeping mats. Also, 20% of orders
include tents.
Let SM denote sleeping mats and let T denote tents.
Determine the following probabilities:
a.
The order includes sleeping mats.
b.
The order includes a tent given it includes sleeping mats.
ANSWERS:
P(SM|T) = 0.4; P(SM|T') = 0.05; P(T) = 0.2
(a) P(SM) = P(SM|T)P(T) + P(SM|T')P(T') = 0.4(0.2) + 0.05(0.8) = 0.12
(b) P(T|SM) =
P(SM|T)P(T)
𝑃(𝑆𝑀)
=
0.4𝑋0.2
0.12
= 0.667
Fradulant
0.0001
T
0.30
0.
𝟕𝟎
0.9999
T
0.01
0.99