I am trying to solve the following question:

If the number of calls received per hour by an answering service is a Poisson random variable with rate of 6 calls per hour, what is the probability of waiting more than 15 minutes between two successive calls?

My understanding of this question is that it's asking what's the probability that there is $0$ call in $15$ min. Therefore, for the Poisson parameter, $\lambda t = 3/2 $ where $t = 0.25$.

Based on the Poisson distribution formula $p(X=x) = \lambda^x \frac{\Large e^{-\lambda}}{x!} $,

the probability that zero call arrives in the next 15 minutes is $p(X = 0) ={(3/2)^0 \frac{\Large e^{-3/2}}{0!}}$ which is ${ e^{-3/2}}$

Is this the correct way of doing it? I have been searching similar questions online and I found someone's answer on reddit using exponential distribution and I dont know if that's correct because the answer is different from mine.

  • 1
    $\begingroup$ Your calculation is correct. The number of calls in $15$ minutes has Poisson distribution parameter $\frac{3}{2}$. The waiting time has exponential distribution parameter $\frac{3}{2}$, which gives an alternative way of doing the problem. Same answer. $\endgroup$ Jul 23, 2015 at 2:42
  • $\begingroup$ Thanks for letting me know! I found the reddit answer and I thought my understand of it was wrong. $\endgroup$ Jul 23, 2015 at 2:45
  • $\begingroup$ You are welcome. It will be important to internalize the relationship between the Poisson and the exponential, but your handling of this problem is perfectly fine. $\endgroup$ Jul 23, 2015 at 2:49
  • $\begingroup$ I looked at your reddit link. Admittedly a small sample, but on that basis a place to avoid for math. $\endgroup$ Jul 23, 2015 at 4:23

1 Answer 1


Your answer with the Poisson distribution is correct. For completeness the anwer with the exponential distribution.

Let $X\sim Exp(\lambda$) with the information that that 6 calls arrived in one hour. The we expect the first call in 10 minutes and we have $$10 = \mathrm{E}[X]=\frac{1}{\lambda}\Rightarrow \lambda = \frac{1}{10}$$

So the probability of waiting more than 15 minutes is

$$P(X>15) = 1- P(X\leq 15) = 1-\left(1-\mathrm{e}^{-15\lambda}\right) = 1-\left(1-\mathrm{e}^{-\frac{15}{10}}\right)=\mathrm{e}^{-\frac{3}{2}}$$

  • $\begingroup$ can you elaborate a bit why in the first formula $\lambda$ is$\frac {1}{10}$? $\endgroup$ Jul 23, 2015 at 23:31
  • $\begingroup$ There are 6 calls per hour and logically there must be 1 call per 10 minutes expected. $\endgroup$
    – Belphegor
    Jul 27, 2015 at 21:19

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