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?
[Math] exponential bank waiting times
probabilityprobability distributions
Best Answer
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, $$ {\rm E[\max \{ X_1 ,X_2 \} ]} = \int_0^\infty {[e^{ - \lambda _1 x} + e^{ - \lambda _2 x} - e^{ - (\lambda _1 + \lambda _2 )x} ]\,dx} = \frac{1}{{\lambda _1 }} + \frac{1}{{\lambda _2 }} - \frac{1}{{\lambda _1 + \lambda _2 }}. $$