Exponential Distribution - memoryless A post office has 2 clerks. Alice enters the post office while 2 other customers,
Bob and Claire, are being served by the 2 clerks. She is next in line. Assume
that the time a clerk spends serving a customer has the Exponential(x)
distribution.
(a) What is the probability that Alice is the last of the 3 customers to be done
being served Hint: no integrals are needed.
(b) What is the expected total time that Alice needs to spend at the post
office?
I understand that no matter who leaves the office first, the remainer has the same expo(x) like Alice, but don't have intuitive thinking why the answer for the 1st is 1/2.
 A: After a while one of $B$ or $C$ is through, say $C$. Now $A$ goes up to the free clerk.
By the memorylessness of the exponential, the additional waiting time for $B$, no matter how long she has already spent being served, has the same exponential distribution as if $B$ had just started being served. So the probability that $B$ finishes before $A$ is $1/2$.
Remark: This is somewhat counterintuitive. And rightly so, because real waiting times do not have exponential distribution. But often the exponential model fits reasonably well.
Because these things can be difficult to see, we look at a discrete analogue of the problem. The discrete analogue of an exponentially distributed random variable is a random variabble with geometric distributions. 
There are two $20$-sided dice, with faces numbered $1$ to $20$.  Customers $B$ and $C$ each are flipping their die, once a minute.  Whenever one of them  gets a $1$, she can go home.
Now $A$ walks in. When a die becomes free, she has to toss it until she gets a $1$. What is the probability $A$ is last to go home?  
This is a bit more complicated, because the two dice could become free at the same time. But that's fairly unlikely.  If only one die becomes free first, say the one used by $C$, then $A$ and $B$ are "equal." The time $B$ has spent fruitlessly tossing doesn't help her go home before $A$. So the probability $A$ is last to go home is about $1/2$. A little more, because of the possibility $B$ and $C$ got their $1$'s simultaneously.   

For (b), we need to know the parameter $\lambda$ of the exponential.  The expected time $A$ spends is the sum of the expected time $A$ waits before a clerk is free, plus the expected time for $A$ to be processed once a clerk is free. The latter is $\frac{1}{\lambda}$.  Now you can look for an argument, intuitive or more formal, about the expected time $A$ waits. For the more formal, note that the probability that neither $B$ nor $C$ is finished after time $t$ is $(e^{-\lambda t})(e^{-\lambda t})$.  
A: Re (a): At some point, either Bob or Clare will leave.  At that time, the clerks will be serving Alice and Bob/Clare.  Because of the memoryless property of the exponential, the expected time for Alice and Bob/Clare is equal, so there is a .5 chance that Alice will finish first and a .5 chance that Bob/Clare will.
A: I have not seen an answer for (b) so I will provide one here. So Alice will get served by the clerk that was serving the first person that leaves, i.e the person that takes the least amount of time between Bob and Claire. Denote $T_i=$ Take that a clerk spends seving costumer $i$. Notice $$L=\min(T_B,Tc) \implies L \sim Expo(\lambda+\lambda)$$
Now, since $$W=\text{Total time spent by Alice}=\min(T_B,Tc)+ T_A$$
we obtain:
$$E(W)=\frac{1}{2\lambda}+\frac{1}{\lambda}=\frac{3}{2\lambda}$$
Only needed to change the final answer since it was added up wrong first
