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Stat 250.3
October 8, 2003
HOMEWORK 3 – SOLUTIONS
7.36
a. P(A) = .55; P(Ac) = .45; P(B|A) = .80; P(B|Ac) = .10.
b. P(A and B) = P(A)P(B|A) = (.55)(.80) = .44. This is the probability of being a Republican and voting for
Candidate X.
c. P(Ac and B) = P(Ac) P(B|Ac) = (.45)(.10) = .045. This is the probability of being a non-Republican and voting for
Candidate X.
d. P(B) = P(A and B) + P(Ac and B) = .485.
e. Candidate X received 48.5% of the votes.
8.10
The sample space of all possible sequences is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
a. For each sequence in the sample space, count the number of tails. Then count how many outcomes there are for
each possible number of tails. Divide by 8 because all outcomes in the sample space are equally likely. The
distribution (pdf) is:
k
P(X =k)
0
1/8
1
3/8
2
3/8
3
1/8
b. For each sequence in the sample space, take the difference (Y) between the number of heads and the number of
tails. For instance, for HHT, Y = 2 (heads)  1 (tails) = 1. List all possible differences, count how many sequences in
the sample space have each possible value, then divide by 8 to get the probability of that value. The distribution (pdf)
is the first one if we care about the sign of the difference, or the second if we do not:
3
1/8
k
P(Y =k)
1
3/8
1
3/8
3
1/8
k
P(Y =k)
1
6/8
3
2/8
c. The sum of heads and tails is 3 for every sequence which means P(Z=3) = 1 so the distribution (pdf) is:
k
P(Z =k)
8.28
a.  = E(X) =
3
1
 xp(x) = (15.8) + (20.2) = 16 minutes.
b. No, the expected value will never be your actual commute time (which is always either 15 or 20 min.)
8.37
a. The number of trials is specified in advance. There are two possible outcomes—either the subject guesses correctly
or not. If the subject merely guesses, the probability of success remains the same from trial to trial. Whether a subject
guesses correctly or not on a trial is independent from the results of previous trials.
b. Yes, X is a binomial random variable with n = 10 and p = .25.
c. The number correct is either 6 or more or 5 or less, so P(X6) =1P(X5) =1.9803 = .0197.
d. With p = .5, P(X6) = 1P(X5) = 1.6230 = .3770.
e. This answer will differ for each student. A factor to consider is that among all people who merely guess,
.0197(about 2%) will be able to get 6 or more correct. If many people are tested, a few who just guess will be able to
get 6 or more right. Another factor to consider is the possible proportion in the population tested that actually has
psychic ability. If few people have psychic ability, a result of 6 or more correct might reasonably be considered to
have been the result of lucky guessing. If many people actually have psychic ability, it might be reasonable to think the
result was obtained from one of those with psychic ability. The discussion of the confusion of the inverse in Section
7.7 is relevant to this discussion.
8.39
a. Note that 1/4 of 1,000 is 250 so the desired probability is P(X250). n = 1000 and p = the proportion of adults in
the United States living with a partner, but not married at the time of the sampling. The value of p is not known.
b. The desired probability is P(X110), n = 500, and p = .20.
c. Note that 70% of 20 is 14 so the desired probability is P(X14). n = 20, and p = .50.
8.40
The formulas are  =np and   np(1  p)
a.  = 10(1/2) = 5 and   10 (.5)(1  .5)  1.5811
b.  = 100(1/4) = 25 and   100 (.25 )(1  .25 )  4.33
c.  = 2500(1/5) = 500 and   2500 (.2)(1  .2)  20
d.  = 1(1/10) = .1 and   1(.1)(1  .1)  .3
e.  = 30(.4) = 12 and   30 (.4)(1  .4)  2.683