Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

156 Some Special Distributions

ith trial. An observed sequence ofnBernoulli trials is then ann-tuple of zeros and
ones. In such a sequence of Bernoulli trials, we are often interested in the total
number of successes and not in the order of their occurrence. If we let the random
variableXequal the number of observed successes innBernoulli trials, the possible
values ofXare 0, 1 , 2 ,...,n.Ifxsuccesses occur, wherex=0, 1 , 2 ,...,n,thenn−x
failures occur. The number of ways of selecting thexpositions for thexsuccesses
in thentrials is (
n
x

)

=

n!

x!(n−x)!

.

Since the trials are independent and the probabilities of success and failure on
each trial are, respectively,pand 1−p, the probability of each of these ways is
p(x(1−p)n−x. Thus the pmf ofX,sayp(x), is the sum of the probabilities of these
n
x

)

mutually exclusive events; that is,

p(x)=

{ (n

x

)

px(1−p)n−x x=0, 1 , 2 ,...,n

0elsewhere.

(3.1.2)

It is clear thatp(x)≥0. To verify thatp(x) sums to 1 over its range, recall the

binomial series, expression (1.3.7) of Chapter 1, which is:

(a+b)n=

∑n

x=0

(

n

x

)

bxan−x,

forna positive integer. Thus,

∑

x

p(x)=

∑n

x=0

(

n

x

)

px(1−p)n−x

=[(1−p)+p]n=1.

Therefore,p(x) satisfies the conditions of being a pmf of a random variableXof
the discrete type. A random variableXthat has a pmf of the form ofp(x)issaid
to have abinomial distribution, and any suchp(x) is called abinomial pmf.A
binomial distribution is denoted by the symbolb(n, p). The constantsnandpare
called theparametersof the binomial distribution.
Example 3.1.1(Computation of Binomial Probabilities).Suppose we roll a fair
six-sided die 3 times. What is the probability of getting exactly 2 sixes? For our
notation, letX be the number of sixes obtained in the 3 rolls. ThenXhas a
binomial distribution withn=3andp=1/6. Hence,

P(X=2)=p(2) =

(

3

2

)(

1

6

) 2 (

5

6

) 1

=0. 06944.

We can do this calculation with a hand calculator. Suppose, though, we want to
determine the probability of at least 16 sixes in 60 rolls. LetYbe the number of
sixes in 60 rolls. Then our desired probability is given by the series

P(Y≥16) =

∑^60

j=16

(

60

j

)(

1

6

)j(

5

6

) 60 −j

,