There are n windows at a ticket office. Currently, n customers are being served at the n windows and k customers are waiting in a single line.

There are n windows at a ticket office. Currently, n customers are being served at the n windows and k customers are waiting in a single line. The amount of time that it takes for a customer to be served at a ticket window is exponentially distributed with parameter L. As soon as a customer at a window is finished being served, the customer leaves the ticket office. The next person in line immediately starts being served at the vacated window. No new customers ever enter the ticket office, and this process continues until all n+k customers have left the ticket office. Let's index the customers. Label the n customers currently being served as 1, …, n. Label the k customers in line as n+1,…, n+k, with the first person in line being labeled n+1 and the last person in line being labeled n+k. Let X be the random variable that is equal to the index of the last customer to leave the ticket office. So for 1 <= i <= n + k, if customer number i is the last person to leave the ticket office, then X = i. If two or more people are simultaneously the last people to leave the ticket office, then we set X = 0. But note that P(X=0)= 0.Now let n and k be arbitrary positive integers. Find P(X=i) for all values 1 <= i <= n + k. That is, for each of the n + k customers, find the probability that customer i is the last to leave the ticket office.