Question

A disk server receives requests from many client machines and requires 8 milliseconds to respond to each request. Let N = the number of additional requests that arrive before the service interval is complete. The probability of exactly k additional requests in the 8-millisecond service interval is P(N = k) = e−0.9(0.9)k/k!, for k = 0,1,2,...,[infinity]. If 2 or more new calls arrive while the service interval is only partially complete, what is the probability that exactly 3 new calls will arrive before the server is ready to respond?

Hint: Use rules for conditional probability.

Answer 1

the probability that exactly 3 new calls will arrive before the server is ready to respond = 0.2165

Explanation:

The concept of conditional probability is applied considering the probability of an event which is about to happen when another event had already occured. The steps and appropriate substitution in the equation given is as shown in the attachment.