Consider a two-server system in which a customer is served first by server $1,$ then by server $2,$ and then departs. The service times at server $i$ are exponential random variables with rates $\mu_{i}, i=1,2 .$ When you arrive, you find server 1 free and two customers at server $2$ customer $A$ in service and customer $B$ waiting in line.
Question: Find $P_{B},$ the probability that $B$ is still in the system when you move over to 2
Instead of using the memoryless property of exponential distribution, and giving some verbal answer to this question, I wanted to show the result by doing explicit calculations; however, I couldn't reach anywhere.
$X_1 :$ the random variable modelling the waiting time of server 1.
$X_2^{k=A,B}:$ the random variable modelling the waiting time of server 2 for each person A and B, separately.
$$P_B = P(X_1 < X_2^{A} + X_2^{B}) = \int_{t=0}^{t=\infty} P\left (t < X_2^{A} + X_2^{B}\right) u_1 e^{-u_1 t} \\ = \int_{t=0}^{t=\infty} \int_{\tau = 0}^{\tau = \infty} P\left (t < \tau + X_2^{B}\right) u_1 e^{-u_1 t} u_2 e^{-u_2 \tau} = \int_{t=0}^{t=\infty} \int_{\tau = 0}^{\tau = \infty} P\left (t - \tau < X_2^{B}\right) u_1 e^{-u_1 t} u_2 e^{-u_2 \tau}\\ = \int_{t=0}^{t=\infty} \int_{\tau = 0}^{\tau = \infty} e^{-u_2 (t-\tau)} u_1 e^{-u_1 t} u_2 e^{-u_2 \tau} = \int_{t=0}^{t=\infty} \int_{\tau = 0}^{\tau = \infty} u_1 u_2 e^{-(u_1+u_2) t} $$ But this integral infinite, since the integrand does not contain any $\tau$.
What is the problem with this "solution"?
$P(t-\tau<X_2^B)\neq e^{-u_2(t-\tau)}$ when $t-\tau<0$. So you need to split the double integral into two cases: $t-\tau\geq 0$ and $t-\tau<0$. That'll give you the dependence on $\tau$.