another follow up question: modeling with exponential distributions This a follow up question to the previous two:  


modeling with exponential distributions  
a follow up question about modeling with exponential distributions




I'm trying to do (c).



Denote the service time for John, Paul and Naomi as $X,Y,Z$ respectively and let $W=X-Y$. Let $T$ be the time Naomi spends in the post office. Then 

$$
\begin{align*}
E(T)&amp;=E(E(T|W))=E(T|W&gt;0)P(W&gt;0)+E(T|W&lt;0)P(W&lt;0)\\
&amp;=E(Z+Y|W&gt;0)P(W&gt;0)+E(Z+X|W&lt;0)P(W&lt;0)\\
&amp;=\left(\frac{1}{\lambda_2}+\frac{1}{\lambda_1+\lambda_2}\right)\frac{\lambda_2}{\lambda_1+\lambda_2}
+\left(\frac{1}{\lambda_1}+\frac{1}{\lambda_1+\lambda_2}\right)\frac{\lambda_1}{\lambda_1+\lambda_2}\\
&amp;=\frac{3}{\lambda_1+\lambda_2}&lt;
\max\left\{1/\lambda_1,1/\lambda_2\right\}
\end{align*}
$$

But I don't see how to get the answer $c=2+\sqrt{3}$.

(Thanks to Did's comment, one should not consider $E(Z)$ as the time Naomi spends in the offices.)
Rewrite the symbols $\lambda_1,\lambda_2$ as $a$ and $b$ respectively. The key point is that we are comparing Naomi's average waiting time to the average waiting time of John and Paul averaged together. 

If one changes $\max\{1/a,1/b\}$ in OP to 
$$
\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)
$$
then one would get
$$
\frac{3}{a+b}&lt;\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)
$$
which would gives the correct answer $c=2+\sqrt{3}$.
Let N, J and P be the service times of Noami, John and Paul.  You have to find the expected time of event N&lt; P-J 

The probability distribution of N = P - J with parameter, $\lambda1$ and $\lambda2$ is a Laplace distribution  with pdf

$f_{N}(n) = \dfrac{e^{\frac{n}{\lambda_2}}}{\lambda_1 + \lambda_2}$ for n&lt;0

$= \dfrac{e^{-\frac{n}{\lambda_1}}}{\lambda_1 + \lambda_2}$ for n>0

Find the expected value of this distribution and using the convention of the problem

$E(N) = \frac{(2\lambda_1 \lambda_2)}{\lambda_1+\lambda_2} $

Let $\lambda_2$>$\lambda_1$.. In other words, max($\lambda_1$,$\lambda_2$) = $\lambda_2$ and min($\lambda_1$,$\lambda_2$) = $\lambda_1$

We also know that $E(N)\lt max(E(P),E(J))$

Using the convention of the problem, where meantime is $\frac{1}{\lambda}$

This implies
$\frac{2\lambda_1\lambda_2}{\lambda_1+\lambda_2}&lt;\frac{1}{\lambda_2}$

$\lambda_2 \gt 2\lambda_1$

In other words, 
max($\lambda_1$,$\lambda_2$) > 2.min($\lambda_1$,$\lambda_2$)

Goes to prove that c = 2

I don't get the number $2+\sqrt{3}$