Probability distribution of number of waiting customers in front of a counter The number of customers arriving at a bank counter is in accordance with a Poisson distribution with mean rate of 5 customers in 3 minutes. Service time at the counter follows exponential distribution with mean service rate of 6 customers in 2 minutes. If there were 20 customers waiting in the line just before opening the counter, what is the probability that more than 20 customers will be waiting in the line, 2 hours after opening the counter? Assume that no customers leave the line without service. 
Here arrival rate is $\lambda=5/3$ and service rate is $\mu=3$.
If $W$ is the average number of people waiting in the line then $W= \lambda/(\mu - \lambda)$ . But I don't know how to calculate number of waiting customer using a given time span. I didn't find any formula to do it in the book. Please help. 
 A: Extended Comment, not Answer.
I do not believe all of the speculations in the Question and Comments are exactly correct and relevant. Here are some things I believe are correct, assuming you are dealing with an M/M/1 queue. 
The formula $W = \lambda/(\mu - \lambda) = \rho/(1 - \rho),$ where $\rho = \lambda/\mu$ is for the average number of customers in an M/M/1 queueing system at equilibrium, including anyone being served. 
The formula $W_Q = \rho^2/(1-\rho)$ is for the average number waiting to be served (again when the system is at equilibrium). 
Such systems reach equilibrium quickly for values of $\lambda$ and
$\mu$ such as yours. Agreeing with @Michael, I think the 2 hours
is supposed to indicate enough time to reach equilibrium. (Markov processes have some memory, but after 'time to equilibrium'
memory of the starting state is assumed irrelevant.) 
The distribution at equilibrium of an M/M/1 queue is well known, and
should be in your text or notes--along with much of what I have said
earlier on. 
A: The 15 minutes mentioned by BrianTung is indeed the exact average time to empty, starting in an initial state of 20 jobs.  Here is the explanation: 
The average duration of a busy period in an $M/M/1$ queue is $\overline{B} = \frac{1}{\mu-\lambda}$.  This can be proven using the renewal theory identity: 
$$ \rho = \frac{\overline{B}}{\overline{B} + \overline{I}} $$
where $\rho=\lambda/\mu$ is the fraction of time busy (with $\rho<1$), and $\overline{I} = 1/\lambda$ is the average idle time in between busy periods. 
The fact that the average time to empty, given an initial state of 20, is just $20\overline{B}$, follows from the very clever (and standard) argument that this time is the same regardless of whether the service is LIFO or FIFO.  With a LIFO (Last-in-First-out) thought experiment, we can imagine each of the 20 initial jobs as generating their own busy periods, each distributed the same as a usual busy period, and these are carried out successively.  Notice that the average time to empty, starting in an initial state of 1 job, is $\overline{B}$. So the total average time to empty is $\overline{B} + \overline{B} + \cdots + \overline{B} = 20\overline{B}$. 
