Advanced High-School Mathematics

SECTION 6.1 Discrete Random Variables 329

Next,

∑∞

n=1

n^2 (1−p)n−^1 =

∑∞

n=1

n(n−1)(1−p)n−^1 +

∑∞

n=1

n(1−p)n−^1

= (1−p)

∑∞

n=1

n(n−1)(1−p)n−^2 +

∑∞

n=1

n(1−p)n−^1

= (1−p)

d^2

dx^2

(1 +x+x^2 +···)

∣∣

∣∣

∣x=1−p+

1

p^2

= (1−p)

d^2

dx^2

(

1

1 −x

)∣∣

∣∣

∣x=1−p+

1

p^2

=

2(1−p)

p^3

+

1

p^2

=

2 −p

p^3

Therefore,

Var(X) =

2 −p

p^2

−

1

p^2

=

1 −p

p^2

6.1.5 The binomial distribution

In this situation we performnindependent trials, where each trial has
two outcomes—call them success and failure. We shall let p be the
probability of success on any trial, so that the probability of failure on
any trial is 1−p. The random variableXis the total number of successes
out of thentrials. This implies, of course, that the distribution of X
is summarized by writing

P(X=k) =

Ñ

n

k

é

pk(1−p)n−k.

The mean and variance ofX are very easily computed once we realize
thatXcan be expressed as a sum ofnindependentBernoullirandom
variables. The Bernoulli random variableBis what models the tossing
of a coin: it has outcomes 0 and 1 with probabilities 1−p and p,
respectively. Very simple calculations shows that

E(B) =p and Var(B) =p(1−p).