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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.

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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 $\operatorname{E}[W_A]$?

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  • $\begingroup$ Many thanks to you :) $\endgroup$ – Vizag May 3 at 22:33
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    $\begingroup$ You say "it is not generally true that $\operatorname{E}[W_A] = \operatorname{E}[S_B]/2$". It does seem to be true here because you have two servers with exponential distributions, so twice the rate and thus half the service time. $\endgroup$ – Henry May 3 at 22:47
  • $\begingroup$ @Henry Yes, but that assertion nevertheless needs justification, which the explicit calculation of the order statistic furnishes. Moreover, my statement remains correct. $\endgroup$ – heropup May 4 at 1:00

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