Page 188 - Fundamentals of Probability and Statistics for Engineers

Some Important Discrete Distributions 171

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
X ˆ X 1 ‡ X 2 ‡ ‡ X r ; 6:26†

where X j is the number of trials between the (j 1)th and (including) the jth
successes. These random variables are mutually independent, each having the
geometric distributionwithmean1/p andvariance (1 p)/p 2 .Therefore,the mean
and variance of sum X are, respectively, according to Equations (4.38) and (4.41),

r 2 r 1 p†
m X ˆ ; ˆ : 6:27†
X
p p 2
Since Y ˆ X r , the corresponding moments of Y are

r 2 r 1 p†
m Y ˆ r; ˆ : 6:28†
Y
p p 2
Example 6.8. Problem: a curbside parking facility has a capacity for three
cars. Determine 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 ˆ 3 and p ˆ 0.8.
Using Equation (6.21), we have

6 6
X X k 1 3 k 3
P X 6†ˆ p k†ˆ 0:8† 0:2†
X
kˆ3 kˆ3 2
3
2
3
ˆ 0:8† ‰1 ‡ 3† 0:2†‡ 6† 0:2† ‡ 10† 0:2† Š
ˆ 0:983:
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

P X 1 r†ˆ P X 2 n†;
6:29†
P X 1 < r†ˆ P X 2 > n†:

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