I have been given a question about solving a M/M/60 (arrival process and service time distributions are Poisson and Exponential respectively) queueing model with $\lambda = 150 $ customers per hour and $\mu = 2$ customers per hour. I am told that the model starts immediately with 100 customers (thus 60 servers are occupied and 40 are placed in the queue, which has infinite capacity).
I am asked to find the probability that a customer immediately arriving after the initial rush (thus is in 41st position in the queue) has a queue wait time of less then $1/3$ of an hour ($20$ minutes).
I understand that 120 customers are expected to be served per hour. Thus the expected time it would take for 40 customers (in the queue) to be served would be 20 minutes. But I need to find the probability that it is less than 20 minutes. I figure that this can be done by finding the probability density function then evaluate the integral between $0 $ and $1/3$ of an hour. But I have no idea how to derive the PDF or if that is even correct.
Any advice is appreciated thanks.
If the service time is $\mathrm{Exp}(\mu)$, then the time for $k$ services has $\mathrm{Erlang}(k,\mu)$ distribution. In this case $\mu=120$ (since there are $60$ servers each with rate $2$) and $k=40$, so the queue wait time of the $101^{\mathrm{st}}$ customer, call it $W$ has density $$ f_W(t) = \frac{(120t)^{39}}{(39)!}120e^{-120t}. $$ It is then straightforward to compute \begin{align} \mathbb P\left(W\leqslant\frac13\right) &= \int_0^{\frac13} f_W(t)\ \mathsf dt\\ &= \int_0^{\frac13} \frac{(120t)^{39}}{(39)!}120e^{-120t}\ \mathsf dt\\ &= 1 - \frac{171341993412178413635784276129625729114158059321}{ 1519760644525099050897380839281}e^{-40}\\ &\approx 0.52109. \end{align}