exponential bank waiting times 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?
 A: For part a), you were almost right. The correct solution is
$$
\frac{1}{{\lambda _1  + \lambda _2 }} + \bigg[\frac{{\lambda _1 }}{{\lambda {}_1 + \lambda _2 }}\frac{1}{{\lambda _1 }} + \frac{{\lambda _2 }}{{\lambda {}_1 + \lambda _2 }}\frac{1}{{\lambda _2 }}\bigg] = \frac{3}{{\lambda _1  + \lambda _2 }} = \frac{3}{{1/3 + 1/6}} = 6,
$$
where we have used the following facts. If $X_i$, $i=1,2$, are independent exponential$(\lambda_i)$ random variables (meaning that they have densities $\lambda_i e^{-\lambda_i x}$, $x > 0$), then $U:=\min\{X_1,X_2\}$ is exponential$(\lambda_1+\lambda_2)$ (and hence its mean is $1/(\lambda_1+\lambda_2)$, which corresponds to the first term above), and moreover, $U$ is independent of the random variable $N$ defined by
$N=1$ if $X_1 < X_2$, and $N=2$ if $X_2 \leq X_1$, for which it holds ${\rm P}(N = 1) = \lambda _1 /(\lambda _1  + \lambda _2 )$ and ${\rm P}(N = 2) = \lambda _2 /(\lambda _1  + \lambda _2 )$. For these facts, see this post (parts (a)-(c)).
EDIT:
For part b), consider 
$$
2 + 2 + \frac{2}{3}6 + \frac{1}{3}3 = 9,
$$
or more generally, 
$$
\frac{1}{{\lambda _1  + \lambda _2 }} + \frac{1}{{\lambda _1  + \lambda _2 }} + \frac{{\lambda _1 }}{{\lambda _1  + \lambda _2 }}\frac{1}{{\lambda _2 }} + \frac{{\lambda _2 }}{{\lambda _1  + \lambda _2 }}\frac{1}{{\lambda _1 }} = \frac{{2\lambda _1 \lambda _2  + \lambda _1^2  + \lambda _2^2 }}{{(\lambda _1  + \lambda _2 )\lambda _1 \lambda _2 }} = \frac{{\lambda _1  + \lambda _2 }}{{\lambda _1 \lambda _2 }}.
$$
(Setting $\lambda_1 = 1/3$ and $\lambda_2 = 1/6$ gives the desired answer, $9$.)
Apparently, you were supposed to solve part b) using the above method.  Nevertheless, it may be worth giving here the following alternative derivation:
$$
\frac{1}{{\lambda _1  + \lambda _2 }} + {\rm E[\max \{ X_1 ,X_2 \} ]} = \frac{1}{{\lambda _1  + \lambda _2 }} + \bigg[
\frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} - \frac{1}{{\lambda _1  + \lambda _2 }}\bigg] = \frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} = 9.
$$
The expression for ${\rm E[\max \{ X_1 ,X_2 \} ]}$ can be derived as follows. First note that
$$
{\rm E[\max \{ X_1 ,X_2 \} ]} =  \int_0^\infty  {{\rm P}(\max \{ X_1 ,X_2 \}  > x)\,dx}  = \int_0^\infty  {[1 - {\rm P}(\max \{ X_1 ,X_2 \}  \le x)} ]\,dx.
$$
Now, using the independence of $X_1$ and $X_2$,
$$
{\rm P}(\max \{ X_1 ,X_2 \}  \le x) = {\rm P}(X_1  \le x){\rm P}(X_2  \le x) = (1 - e^{ - \lambda _1 x} )(1 - e^{ - \lambda _2 x} ),
$$
and hence
$$
1 - {\rm P}(\max \{ X_1 ,X_2 \}  \le x) = e^{ - \lambda _1 x}  + e^{ - \lambda _2 x}  - e^{ - (\lambda _1  + \lambda _2 )x} .
$$
Finally, 
$$
{\rm E[\max \{ X_1 ,X_2 \} ]} = \int_0^\infty  {[e^{ - \lambda _1 x}  + e^{ - \lambda _2 x}  - e^{ - (\lambda _1  + \lambda _2 )x} ]\,dx}  = \frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} - \frac{1}{{\lambda _1  + \lambda _2 }}.
$$
A: 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 $\lambda_2 = 1/6$).
