# Poisson process with time exponentially distributed

While studying Poisson processes, I have found a problem I can't solve:

Two people, A and B, arrive at a bank and wait to be attended. A and B are in the waiting queue and A is before B. The workers attend the customers according to a Poisson process of parameter λw. A and B are willing to wait, respectively, a time TA and TB before going away. TA and TB are exponentially distributed with parameters λA and λB.

What is the probability that the person B is attended before he or she leaves?

How can we calculate the answer?

• What have you tried? WIthout showing some work, and saying where you got stuck, it just looks like you are trying to get somebody to solve a homework problem for you. – Mark Fischler May 22 at 19:51

Assuming that the patience times apply equally whether a customer is in service or waiting for service, the time customer A is in the system is exponentially distributed with rate $$\lambda_A+\lambda_W$$. Hence the probability that customer B is attended is $$\frac{\lambda_A+\lambda_W}{\lambda_A+\lambda_W+\lambda_B}.$$