Economics 70 Answer key to Book Problem 3
Exercise Title: Book Problem 3
Date: 5/12/2009
Name (Last, First): ANSWER KEY
PID:
Section No. of Course: Sections 1 & 2
Sign the Honor Pledge:
3.17
a)
The probability that a person is a college graduate, P(A), is 0.24.
The
probability that a
college graduate believes the ad, P(BA), is 0.18, so the probability that a college grad
does not believe the ad, P(
B
A ), is 1 – 0.18 = 0.82.
Thus, the P(
B
 A)P(A) =
(0.82)*(0.24) = 0.197
b)
The probability that an adult with some college believes the ad is 0.25, so the probability
that an adult with some college does not believe the ad is 1 – 0.25 = 0.75
c)
The probability that an adult has not been to college, P(C), is 0.4.
The probability that
someone who has not been to college believes the ad, P(D), is 0.27.
The joint
probability, P(CD) = P(D  C)P(C) = (0.27)*(0.4) = 0.108
3.19
Define the following events:
A:
car has been in an accident in the past year
B:
car has antilock brakes
a)
P(A) = P(AB) + P(A
B
) = 0.03 + 0.12 = 0.15
b)
P(ĀB) = 0.40
c)
P(B  A) = P(AB)
= 0.03/0.15 = 0.2
P(A)
3.21
Define the event A: executive will use the information superhighway.
a)
P(A) = 0.40
b)
If two executives are randomly chosen, a simple event consists of a pair, the first element
representing the first executive’s action, and the second representing the action of the
second executive.
Then the event that only one of the two executives will use the
Information Superhighway is composed of two simple events, AĀ and ĀA.
The
probability is
P(AĀ) + P(ĀA) = (0.4)*(0.6) + (0.6)*(0.4) = 0.48
3.23
Define the following events:
A: Device A is activated
B: Device B is activated
It is given that P(A) = 0.91 and P(B) = 0.95, and that the two systems (and hence, events
A and B) are independent.
Then
P(system functions) = P(either or both device activates)
1
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= P(A
⋃
B) = P(A) + P(B) – P(AB)
= P(A) + P(B) – P(A)P(B) = 0.91 + 0.95 – (0.91)*(0.95) = 0.9955
3.25
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 Spring '08
 turchi
 Economics, Probability, Probability distribution, Histogram, Probability space

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