I have following problem to solve :

There are $N$ customers waiting in the checkout line, where $N$ has Poisson distribution with parameter $3$. If a service begins when $m$ customers are in the line, that service time has exponential distribution with parameter $\frac{1}{2m}$ . Find the probability that a customer just approaching the checkout will be served in less than a minute.

What I did :

If X is a time serving, then


according to CDF for exponential distribution. But now I don't know what to do, how to use fact that $N$ has Poisson distribution. I'm puzzled. How to connect these two things?


1 Answer 1


Use conditional probability. Let $$N\sim \text{Poisson}(3)\\ X|N\sim \text{Exponential}(1/(2N))$$

Then from total probability $$P(X<1)=\sum_{N=0}^\infty F(1|N)P(N)$$

$$\begin{split}\sum_{N=0}^\infty (1-e^{-\frac 1{2N}})\frac{e^{-3}3^N}{N!}&=\sum_{N=0}^\infty \frac{e^{-3}3^N}{N!}-e^{-3}\sum_{N=0}^\infty \frac{e^{-\frac 1{2N}}3^N}{N!}\\ &=1-???\end{split}$$

  • $\begingroup$ thanks for your answer. I did it and got $P(X<1)=F(1|1) \cdot P(1)+F(1|2) \cdot P(2)+...= (1-e^{-\frac{1}{2}}) \cdot 3e^{-3} +(1-e^{-\frac{1}{4}}) \cdot \frac{9e^{-3}}{2} + ... $ and I don't know what's next, how to calculate it till the end. Does it even converge? $\endgroup$
    – FermatFan
    Jun 10, 2021 at 16:51
  • $\begingroup$ @FermatFan good point. I have updated the answer but like yourself am not sure how to finish it. $\endgroup$
    – Vons
    Jun 10, 2021 at 17:32
  • $\begingroup$ Wolfram shows that it converges and sum equals approximately to 15.5, but I have no idea how is it calculated. $\endgroup$
    – FermatFan
    Jun 10, 2021 at 17:55
  • $\begingroup$ @Stacker : the series goes from 1 and not from 0. Calculating it manually ($\sum_1^9$ is enough) I got $p\approx 17.82\%$ $\endgroup$
    – tommik
    Jun 12, 2021 at 6:18

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