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.

  • $\begingroup$ Note that there are infinitely many time windows in which a call could be answered, not just $20$, because there is no stipulation that calls have to answered at exactly three-minute intervals. There are general solutions to this problem, which is a simple instance of those studied by queuing theory. $\endgroup$
    – whuber
    Dec 19, 2016 at 18:56
  • $\begingroup$ @whuber - Breaking the hour down into equal intervals was intended as a simplification of the problem, but based on your comment I'm no longer sure that it is actually a simplification. I'm open to treating time as continuous rather than binning if it makes thing easier. $\endgroup$
    – dannyhmg
    Dec 19, 2016 at 19:02
  • $\begingroup$ Whether it's a simplification or not, it could change the answers (although probably not too much). It could also help you to clarify some of the concepts. For instance, what does "simultaneously" mean? It probably differs from the usual sense of that word: I suspect in this context it ought to mean "none of the five calls occur while the receptionist is handling a call." $\endgroup$
    – whuber
    Dec 19, 2016 at 20:21

1 Answer 1


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 $\rho{W}$.

  • $\begingroup$ I had worked out a solution based on @whuber's feedback and some research into queuing theory, and I was in the process of adding my solution to the question when you posted your answer. It looks to me like your answer and my solution are equivalent (which makes me happy), but please let me know if I missed something. Thanks! $\endgroup$
    – dannyhmg
    Dec 19, 2016 at 21:47
  • $\begingroup$ Say call 1 comes in at $t_1=0$, so it will finish at $t=3$. Then call 2 comes in at $t_2=1<3$, so it must be put in the queue, and will be completed at $t=3+3=6$. Now if call 3 comes in at $t_3=5<6$, it must be put into the queue even though $t_3-t_2=4>3$. $\endgroup$
    – GeoMatt22
    Dec 19, 2016 at 22:09
  • $\begingroup$ That's a good point. It seems to me that the binomial approximation is reasonable as long as the calls are not too frequent, but breaks down when calls start to pile on top of each other (for the reason you mentioned). $\endgroup$
    – dannyhmg
    Dec 20, 2016 at 11:46
  • $\begingroup$ The binomial solution is exact if all missed calls are dropped, i.e. if there is no queue (so an M/D/1/0 queue). $\endgroup$
    – GeoMatt22
    Dec 20, 2016 at 18:33

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