Advanced High-School Mathematics

SECTION 6.1 Discrete Random Variables 331

E(X) =

r

p

, Var(X) =

r(1−p)

p^2

.

The name “inverse binomial” would perhaps be more apt, as the
following direct comparison with the binomial distribution reveals:

Binomial Random Variable Negative Binomial Random Variable

X Y

number of successes number of trials

inntrials needed forrsuccesses

E(X) =np, Var(X) =np(1−p) E(Y) =

r

p

, Var(Y) =

r(1−p)

p^2

Generalization 2: The coupon problem

Suppose that in every cereal box there is a “prize,” and that there are,
in all, three possible prizes. Assume that in a randomly purchased
box of cereal the probability of winning any one of the prizes is the
same, namely 1/3. How many boxes of cereal would you expect to buy
in order to have won all three prizes? It turns out that the natural
analysis is to use a sum of geometric random variables.
We start by defining three independent random variables X 1 , X 2 ,
andX 3 , as follows. X 1 is the number of trials to get the first new prize;
note thatX 1 is really not random, as the only possible value of X 1
is 1. Nonetheless, we may regardX 1 as a geometric random variable
with probabilityp= 1. X 2 is the number of trials (boxes) needed to
purchase in order to get the second new prize, after the first prize is
already won. ClearlyX 2 is also a geometric random variable, this time
withp= 2/3. Finally,X 3 is the number of boxes needed to purchase to
get the third (last) new prize after we already have two distinct prizes,
and soX 3 is geometric withp= 1/3. Therefore, ifX is the number of
boxes purchased before collecting the complete set of three prizes, then
X=X 1 +X 2 +X 3 , which representsXas a sum of geometric random
variables.
From the above, computingE(X) is now routine: