Probability of N bins containing events for a poisson process Suppose a receptionist, answering phones at a business, takes 3 minutes to help each caller. Thus, in each hour, there are 20 time windows in which a call could be answered.  Calls arrive following a poisson process with an average rate of 5 calls per hour.  If 5 calls arrive during a particular hour, what is the probability that none of the 5 calls occur while the receptionist is busy, so that no calls are missed?  What is the probability of exactly 1 missed call?  Of 2 missed calls?  Is there a general solution to this kind of problem given N calls and W windows of time?  
Here's the solution I came up with:   
The arrival of calls is a Poison process, so the probability of a gap between calls of 3 minutes or more is p = $e^{-\lambda t} = e^{-\lambda * 3} $. Now if we have $N = 5$ calls in an hour, then there are $N-1 = 4$ gaps and the probability that all of the gaps are 3 minutes or longer is  
P(no missed calls) = $ (e^{-\lambda * 3})^{N-1} = (e^{-\lambda * 3})^4$  
The probability of exactly one missed call is the probability that one and only one gap between calls is shorter than 3 minutes: 
P(1 missed call) = $ 4* (e^{-\lambda * 3})^{4-1} * (1-e^{-\lambda * 3})^1 $
Where the pre-factor of 4 comes from the fact that any of the 4 gaps can be the short one.  Similarly, the probability of exactly two missed calls is
P(2 missed call) = $ (^4_2)*(e^{-\lambda * 3})^{4-2} * (1-e^{-\lambda * 3})^2 $
Clearly this generalizes to a binomial distribution where the probability p (small p) is a function of the arrival rate $\lambda$. 
I'd be grateful for any feedback as to whether this is correct/reasonable.
 A: For reference, this would be called an "M/D/1 queue" in queuing theory.
The parameters for this model are the arrival rate $\lambda$ (= 5 calls/hour) and the service rate $\mu$ (= 20 calls/hr), which combine to give the utilization $\rho=\lambda/\mu$ (= 1/4).
For a Poisson process with rate parameter $\lambda$ the time between arrivals will follow an exponential distribution with mean $1/\lambda$. This means the probability that no new call arrives while a given call is being serviced is
$$\Pr\big[t>\tfrac{1}{\mu}\big]=e^{-\rho}$$
which in your case gives $\Pr[t>3\text{ minutes}]\approx{77.9\%}$.
Due to the memoryless property of the inter-arrival times, the probability of a given call being interrupted should be independent of the time between that call and the previous one. So the probability of no interruptions over $N$ calls should just be $p=e^{-(N-1)\rho}\approx{36.8\%}$ in your $N=5$ case.

The probability of getting $k$ uninterrupted calls out of $N$ calls in time $T$ is more complicated. A simple approximation could possibly be to assume $k$ follows a Binomial distribution with success probability $p$ and $N$ trials?
Note: This is definitely not correct, as it does not account for queuing time. For example if there is an interruption in one call at $t<T_0=\frac{1}{\mu}$, then this call will have to wait a time $T_0-t$ before being answered. So a third call will have to wait if it comes in within time $2T_0-t$ of the second call.

I did not use your $W$ parameter here at all, but the number of calls $N$ occurring over a time $\frac{W}{\mu}$ will be given by a Poisson distribution with parameter $\rho{W}$. 
