Question

Customers arrive at a service center according to a Poisson process with a mean interarrival

time of 15 minutes.

1. If two customers were observed to have arrived in the first hour, what is the probability that at least

   one arrived in the last 10 minutes of that hour?

Answer 1

Customers arrive at a mean interarrival rate of 15 minutes.

Thus in an hour mean of number of customers = 60/15 = 4.  

The average number of customers in 10 min = 10/15 = 2/3

The average number of customers in 50 min = 50/15 = 10/3

We have the following two cases:

Case1: Both customers arrive in the last 10 min of that hour and none arrive in the first 50 min.

According to poisson process,

Case 2: One customer comes in the first 50 min of the hour and the second customer comes in last 10 min of the hour

Thus the probability that only one of the two persons arrives in last 10 min of the hour,

P(X=1) =

Now, probability that two customers arrive in the first hour =

Thus according to Bayes's formula, that probability that at least one customer arrived in the last 10 min

= P(X=1) + P(X=2) =   = 0.3056