TITLE: exponential bank waiting times
QUESTION [6 upvotes]: I have a question about expected waiting times at a bank.
Consider a bank with two tellers. Three people, A, B and C enter the bank at almost the same time and in that order. A and B go directly into service while C waits for the first avaliable teller. Suppose that the service time for teller one is exponentially distributed with mean 3 and teller two with mean 6.
a) What is the expected total amount of time for C to complete her business?
b) What is the expected total time until the last of the three customers leaves?
c) What is the probability C is the last one to leave?

REPLY [2 votes]: Here is a tricky way to solve part b). (I decided to put this in a new answer, since the first one is too long already.)
The expected total time for part b) is given by
$$
E = {\rm E}[\min \{ X_1 ,X_2 \} ] + {\rm E}[\max \{ \tilde X_1 ,\tilde X_2 \} ],
$$
where $X_i$, $i=1,2$, are independent exponential$(\lambda_i)$ random variables and $(\tilde X_1,\tilde X_2)$ is an independent copy of $(X_1,X_2)$. Hence 
$$
E = {\rm E}[\min \{ X_1 ,X_2 \} ] + {\rm E}[\max \{ X_1 ,X_2 \} ] = {\rm E}[\min \{ X_1 ,X_2 \} + \max \{ X_1 ,X_2 \}].
$$
But $\min \{ X_1 ,X_2 \} + \max \{ X_1 ,X_2 \} = X_1 + X_2$, and hence
$$
E = {\rm E}[X_1 + X_2] = \frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }}
$$
(which is equal to $9$ if $\lambda_1 = 1/3$ and $\lambda_2 = 1/6$).