Expected Waiting Time in a Queue with exponential distribution I am trying to solve the following problem in my exercise sheet towards preparation of competitive exam: 
A post office has 2 clerks. $A$ enters the post office while 2 other customers, $B$ and $C$, are being served by the 2 clerks. She is next in line. Assume that the time a clerk spends serving a customer has the Expo($\lambda$) distribution.
What is the expected time spent by $A$ in the post office?
My take: 
Total time that A spends in the office is the sum of waiting time + the time taken to serve A. 
$$E(\text{Total Time}) = E(\text{Waiting Time}) + E(\text{Service Time})$$
$$E(\text{Service Time}) = \frac{1}{\lambda}$$
Now I computed that the Probability that A is served at any of the counter is $1/2$. Therefore,
$$E(\text{Waiting Time}) = 1/2 \times E(\text{Service Time of B}) + 1/2 \times E(\text{Service Time of C})$$
$$=\frac{1}{\lambda}$$
Which gives me 
$$E(\text{Total Time}) = \frac{2}{\lambda}$$
But the answer given is $\frac{3}{2\lambda}$.
Please point out the mistake in what I am thinking or if you think there's a better way to do it.
Thanks.
 A: The mistake is that the waiting time is not distributed in the way that you have proposed, because the waiting time is the lesser of the service times of $B$ and $C$.  This is because $A$ takes the first available opening, so whoever finishes first, $A$ chooses that clerk.  That is to say, if $W_A$ is the waiting time of $A$ and $S_B$, $S_C$ are the service times of $B$ and $C$ respectively, then $$W_A = \min(S_B, S_C),$$ and it is not generally true that $\operatorname{E}[W_A] = \operatorname{E}[S_B]/2$.
Rather, you must compute the minimum order statistic of $B$ and $C$'s service times.  Since these are iid exponentially distributed with rate $\lambda$, it follows that $$\Pr[W_A > t] = \Pr[\min(S_B, S_C) > t] = \Pr[(S_B > t) \cap (S_C > t)] \overset{\text{ind}}{=} \Pr[S_B > t] \Pr[S_C > t] \overset{\text{id}}{=} (e^{-\lambda t})^2.$$  That is to say, the probability that $A$ waits more than $t$ to begin service is the product of the probabilities that both $B$ and $C$ take more than $t$ to be finished.  And this makes perfect intuitive sense--because if $A$ must wait more than $t$, then that means both $B$ and $C$ are still occupied, which means their service times are also more than $t$.
Now that you know $\Pr[W_A > t] = e^{-2\lambda t}$, what can you say about $\operatorname{E}[W_A]$?
