Chapter 3 Problems
1. Let Rj,Mj, and Ajrespectively designate the events, {red on jth draw},
{mauve on jth draw}, and {attractive rainbox motif on jth draw}. Be-
cause there are 4 red socks, 6 mauve ones, and 8 rainbow-colored socks(!),
we have
Pr{match}= Pr(R1∩R2) + Pr(M1∩M2) + Pr(A1∩A2)
=4
2
18
2+6
2
18
2+8
2
18
2=49
153.
5. Worked out during lectures.
27. Let p= Pr(H), where H={heads}, etc. Note that the probability of the
event A={number of heads = the number of tails}is the same as the
probability that the total number of heads is exactly n. Let Hdenote the
collection of all arrangements of nheads and ntails [try to digest this
first!]. Each B∈ H has probability pn(1 −p)n. Therefore,
Pr(A) = |H| · pn(1 −p)n=|H|(pq)n,
where |H| denotes the number of elements of Hand q= 1 −p, as usual.
It should now be clear that |H| is the number of ways we can distribute n
heads—and hence also ntails—in 2nslots. Thus, |H| =2n
n. Therefore,
Pr(A) = 2n
n(pq)n.(1)
This is the desired formula. In order to approximate this expression for
large values of n, we need the Stirling formula (p. 96 of your text):
n!∼(n/e)n√2πn, (2)
where an∼bnmeans limn→∞(an/bn) = 1.
We apply (2) to find that
2n
n=(2n)!
(n!)2∼(2n/e)2n√4πn
(n/e)n√2πn2=(2n/e)2n√4πn
2πn ·(n/e)2n=22n
√πn
=4n
√πn .
Plug this into (1) to find that
Pr(A)∼(4pq)n
√πn as n→ ∞.
An interesting feature of this is that when p=q= 1/2, then Pr(A)∼
1/√πn goes to zero as n→ ∞, but rather slowly. On the other hand,
1
