Jim Lambers
Math 6A
Spring Quarter 2003-04
Lecture 15 Examples
This examples corresponds to Section 5.3 in the text.
Example (Exercise 25, Section 5.3) Let Xnbe the random variable that counts the difference in
the number of tails and the number of heads when ncoins are flipped. What is the expected value
of Xn, and what is its variance?
Solution This problem is similar to one given in Quiz 5, except that a specific value of nwas
given. For the case of an arbitrary n, we must first determine the possible values of Xn. For
j= 0,1,2, . . . , n, if the number of tails in a sequence of nflips is j, then the number of heads in
the same sequence is n−j. It follows that the difference in the number of tails and the number of
heads is j−(n−j) = 2j−n. Therefore, the possible values of Xnare 2j−n, for j= 0,1,2, . . . , n.
In order to compute the expected value and the variance, we must know the probability of j
tails occurring, for each j= 0,1,2, . . . , n. To compute this probability, we can use the fact that
each coin flip is an independent Bernoulli trial with probability of success p= 1/2, where success
is defined to be a tail and the coin is assumed to be fair. It follows that for each j= 0,1,2, . . . , n,
the probability of jtails is C(n, j)(1/2)j(1/2)n−j= 2−nC(n, j).
The probability of jtails is the same as the probability that Xn= 2j−n, so using the formula
E(X) = X
r∈X(Sn)
rp(X=r),
where Snis the sample space of all sequences of ncoin flips, we obtain
E(Xn) =
n
X
j=0
(2j−n)2−nC(n, j)
= 2−n
n
X
j=0
(2j−n)C(n, j)
= 2−n
n
X
j=0
2jC (n, j)−
n
X
j=0
nC(n, j )
= 2−n
2
n
X
j=1
jC (n, j)−n
n
X
j=0
C(n, j)
= 2−n
2
n
X
j=1
jC (n, j)−n2n
1
