Question

The time of arrival at a toll booth on a certain road comes from a Poisson process with an average of 9 vehicles per minute. Now, X is the time that is measured until the arrival of the 5th vehicle. a) What is the probability that X is at least half a minute? b) What is the average time of X?
Ans. a) 0.532104, b) 5/9

Answer 1

a)

We know that the inter-arrival time of Poisson process follows exponential distribution. Let T1, T2,.., T5 be the inter-arrival time of 1st , 2nd,..., 5th vehicle. Then Ti ~ Exp( = 1/9) for   = 1,2,3,4,5

X = T1 + T2 + T3 + T4 + T5

We know that the 'n' sum of exponential distribution follows Gamma distribution with parameters n and

X ~ Gamma(k = 5, = 1/9)

Probability that X is at least half a minute = P(X 0.5)

(CDF of Gamma distribution)

= 0.011109 * (1 + 4.5 + 4.5² / 2 + 4.5³ / 6 + 4.5⁴ / 24 )

= 0.532104

b)

Average time of X = E[X] = k = 5 * (1/9) = 5/9