the first subsequent candidate who is best among
those already interviewed (if no such candidate
appears, the last one interviewed is hired).
For example, with s ¼2, the order 3, 4, 1,
2 would result in the best being hired, whereas
the order 3, 1, 2, 4 would not. Of the four possible
s values (0, 1, 2, and 3), which one maximizes
P(best is hired)? [Hint: Write out the 24 equally
likely interview orderings: s ¼0 means that the
first candidate is automatically hired.]
104. Consider four independent events A
1
, A
2
, A
3
,
and A
4
and let p
i
¼P(A
i
)fori ¼1, 2, 3, 4.
Express the probability that at least one of
these four events occurs in terms of the p
i
’s,
and do the sam e for the probability that at least
two of the events oc cur.
105. A box contains the following four slips of paper,
each having exactly the same dimensions: (1) win
prize 1; (2) win prize 2; (3) win prize 3; (4) win
prizes 1, 2, and 3. One slip will be randomly
selected. Let A
1
¼{win prize 1}, A
2
¼{win
prize 2}, and A
3
¼{win prize 3}. Show that A
1
and A
2
are independent, that A
1
and A
3
are inde-
pendent, and that A
2
and A
3
are also independent
(this is pairwise independence). However, show
that P(A
1
\A
2
\A
3
) 6¼P(A
1
) · P(A
2
) · P(A
3
), so the
three events are not mutually independent.
106. Consider a woman whose brother is afflicted
with hemophilia, which implies that the woman’s
mother has the hemophilia gene on one of her
two X chromosomes (almost surely not both,
since that is generally fatal). Thus there is a
50–50 chance that the woman’s mother has
passed on the bad gene to her. The woman has
two sons, each of whom will independently
inherit the gene from one of her two chromo-
somes. If the woman herself has a bad gene, there
is a 50–50 chance she will pass this on to a son.
Suppose that neither of her two sons is afflicted
with hemophilia. What then is the probability
that the woman is indeed the carrier of the hemo-
philia gene? What is this probability if she has a
third son who is also not afflicted?
107. Jurors may be a priori biased for or against the
prosecution in a criminal trial. Each juror is
questioned by both the prosecution and the
defense (the voir dire process), but this may not
reveal bias. Even if bias is revealed, the judge
may not excuse the juror for cause because of the
narrow legal definition of bias. For a randomly
selected candidate for the jury, define events B
0
,
B
1
, and B
2
as the juror being unbiased, biased
against the prosecution, and biased against the
defense, respectively. Also let C be the event that
bias is revealed during the questioning and D be
the event that the juror is eliminated for cause.
Let b
i
¼P(B
i
)(i ¼0, 1, 2), c ¼P(C|B
1
) ¼P(C|B
2
)
and d ¼P(D|B
1
\ C) ¼P(D|B
2
\ C)[“FairNum-
ber of Peremptory Challenges in Jury Trials,”
J. Amer. Statist. Assoc., 1979: 747–753].
a. If a juror survives the voir dire process, what
is the probability that he/she is unbiased (in
terms of the b
i
’s, c, and d)? What is the prob-
ability that he/she is biased against the prose-
cution? What is the probability that he/she is
biased against the defense? [Hint: Represent
this situation using a tree diagram with three
generations of branches.]
b. What are the probabilities requested in (a) if
b
0
¼.50, b
1
¼.10, b
2
¼.40 (all based on data
relating to the famous trial of the Florida
murderer Ted Bundy), c ¼.85 (corresponding
to the extensive questioning appropriate in a
capital case), and d ¼.7 (a “moderate” judge)?
108. Allan and Beth currently have $2 and $3, respec-
tively. A fair coin is tossed. If the result of the
toss is H, Allan wins $1 from Beth, whereas if the
coin toss results in T, then Beth wins $1 from
Allan. This process is then repeated, with a coin
toss followed by the exchange of $1, until one of
the two players goes broke (one of the two
gamblers is ruined). We wish to determine
a
2
¼P(Allan is the winner j he starts with $2)
To do so, let’s also consider a
i
¼P(Allan wins j
he starts with $i) for i ¼0, 1, 3, 4, and 5.
a. What are the values of a
0
and a
5
?
b. Use the law of total probability to obtain an
equation relating a
2
to a
1
and a
3
.[Hint: Con-
dition on the result of the first coin toss,
realizing that if it is a H, then from that
point Allan starts with $3.]
c. Using the logic described in (b), develop a
system of equations relating a
i
(i ¼1, 2, 3, 4)
to a
i–1
and a
i+1
. Then solve these equations.
[Hint: Write each equation so that a
i
a
i–1
is
on the left hand side. Then use the result of the
first equation to express each other a
i
a
i1
as a function of a
1
, and add together all four of
these expressions (i ¼2, 3, 4, 5).]
d. Generalize the result to the situation in which
Allan’s initial fortune is $a and Beth’s is $b.
Note: The solution is a bit more complicated
if p ¼P(Allan wins $1) 6¼.5.
94
CHAPTER 2 Probability