Fundamentals of Probability and Statistics for Engineers

which is the reason for the name ‘negative binomial distribution’.
The mean and variance of random variable X can be determined either by
following the standard procedure or by noting that X can be represented by

where Xj is the number of trials between the (j 1)th and (including) the jth
successes. These random variables are mutually independent, each having the
geometricdistributionwithmean1 andvariance .Therefore,themean
and variance of sum X are, respectively, according to Equations (4.38) and (4.41),

Since , the corresponding moments of Y are

Example 6.8.Problem: a curbside parking facility has a capacity for three
cars. D etermine the probability that it will be full within 10 minutes. It is
estimated that 6 cars will pass this parking space within the timespan and, on
average, 80% of all cars will want to park there.
Answer: the desired probability is simply the probability that the number of
trials to the third success (taking the parking space) is less than or equal to 6. If
X is this number, it has a negative binomial distribution with r 3 and p 0 .8.
U sing Equation (6.21), we have

Let us note that an alternative way of arriving at this answer is to sum the
probabilities of having 3, 4, 5, and 6 successes in 6 Bernoulli trials using the
binomial distribution. This observation leads to a general relationship between
binomial and negative binomial distributions. Stated in general terms, if X 1 is
B(n, p) and X 2 is NB(r, p), then

Some Important Discrete Distributions 171

XˆX 1 ‡X 2 ‡‡Xr; … 6 : 26 †

/p (1p)/p^2

mXˆ

r

p

;^2 Xˆ

r… 1 p†

p^2

: … 6 : 27 †

YˆXr

mYˆ

r

p

r;^2 Yˆ

r… 1 p†

p^2

: … 6 : 28 †

ˆ ˆ

P…X 6 †ˆ

X^6

kˆ 3

pX…k†ˆ

X^6

kˆ 3

k 1

2



… 0 : 8 †^3 … 0 : 2 †k^3

ˆ… 0 : 8 †^3 ‰ 1 ‡… 3 †… 0 : 2 †‡… 6 †… 0 : 2 †^2 ‡… 10 †… 0 : 2 †^3 Š

ˆ 0 : 983 :

P…X 1 r†ˆP…X 2 n†;

P…X 1 <r†ˆP…X 2 >n†:

… 6 : 29 †