Advanced High-School Mathematics

334 CHAPTER 6 Inferential Statistics

• HH

HH

HHT

(B 1 = 0 and we start the experiment over again

with one trial already having been performed.)





 HH

HHH

T

(B 1 = 1, B 2 = 0 and we start the experiment

over again with two trials already having been

performed.)





H

H (B 1 =B 2 = 1 and the game is over)

1 −p

p

1 −p

p

Computing the expectation of both sides of (6.6) quickly yields

E(X) = 2p^2 +p(1−p)(2 +E(X)) + (1−p)(1 +E(X)),

from which it follows that

E(X) =

1 +p

p^2

.

Note that if the coin is fair, then the expected waiting time before
seeing two heads in a row is 6.

Similar analyses can be applied to computing the expected wait-
ing time before seeing the sequence HT (and similar) sequences, see
Exercises 8, 9, and 10 on page 341.

6.1.7 The hypergeometric distribution

This distribution is modeled by a box containingN marbles, of which
nof these are of a particular type (“successful” marbles) and so there
areN −n “unsuccessful” marbles. If we draw k marbles without re-
placement, and ifXis the random variable which measures the number
of successful marbles drawn, thenX has distribution given by

P(X=m) =

Än

m

äÄN−n

k−m

ä

ÄN

k

ä , m= 0, 1 , 2 ,...,max{n,k}.

From the above it follows that the mean ofXis given by the sum