Exponential Probability of customers Assume there is post office that is run by two clerks $1$ and $2$. Customers arrive at times that follow exponential distribution with rate $λ_a$. The amount of time that the clerks 1 and 2 serve are exponentially distributed with mean $1/λ_1$ and $1/λ_2$. The first customer that arrives in the post office is served by clerk $1$ and the second by clerk $2$. The three exponential random variables are independent. Two customers A and B have decided to go to the post office. What is the probability that A arrives before and departs after B?
So, $T_1$ = time taken by clerk $1$ and $T_2$ = time taken by clerk 2. To solve for the probability, if A is the first one there, she goes to clerk $1$ and clerk $1$ has to be faster than clerk $2$ ($T_1 < T_2$). Now what?
 A: The question of the probability density of $T_0+T_2$ arises:
\begin{align}
f_{T_0+T_2}(t) & = \int_0^t f_{T_0}(s) f_{T_1} (t-s) \, ds = \int_0^t (e^{-\lambda_a s} \cdot\lambda_a)\, e^{-\lambda_2(t-s)} (\lambda_2\,ds) \\[10pt]
& = \lambda_a\lambda_2 e^{-\lambda_2 t} \int_0^t e^{(\lambda_2-\lambda_a)s} \,ds =\lambda_a \lambda_2 e^{-\lambda_2 t} \left[ \frac{e^{(\lambda_2-\lambda_a)s}}{\lambda_2 -\lambda_a} \right]_{s\,:=\,0}^{s\,:=\,t} \\[10pt]
& = \lambda_a\lambda_2 e^{-\lambda_2 t} \cdot \frac{e^{(\lambda_2 - \lambda_a) t} - 1}{\lambda_2 - \lambda_a} = \frac{\lambda_a\lambda_2}{\lambda_2 - \lambda_a} \left( e^{-\lambda_a t} - e^{-\lambda_2 t} \right) \text{ for } t \ge 0.
\end{align}
In case $\lambda_a=\lambda_2,$ one might think that the limit of that last expression as $\lambda_a\to\lambda_2$ would be the thing to find. Maybe what is a bit simpler is that the integral on the second line is easy to evaluate in that case.
Then we have
\begin{align}
\Pr(T_1 < T_0+T_2) & = \operatorname E(\Pr(T_1 < T_0 + T_2) \mid T_0+T_2) = \operatorname E\left( 1 - e^{-\lambda_1(T_0+T_2)}\right) \\[10pt]
& = 1 - \int_0^\infty t f_{T_0+T_2} (t) \, dt 
\end{align}
and so on.
To find this last integral, one should recall that
$$
\int_0^\infty t e^{-ct} \, dt = \frac 1 {c^2} \int_0^\infty (ct) e^{-ct} (c\, dt) = \frac 1 {c^2} \int_0^\infty ue^{-u} \, du = \frac 1 {c^2}.
$$
(One can derive this by integrating by parts $\displaystyle \int u \Big(e^{-u}\,du\Big) = \int u\,dv = \cdots$
