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Suppose there is a shop with one counter and follows first-come-first-service rule. The arrival rate of customers follows Poisson distribution while the expected service time follows exponential distribution. It is given that, $2$ customers arrive every $10$ minutes while $18$ customers are served per hour.

What is the probability that an incoming customer to wait for more than $30$ minutes before being served ?


The problem involves two distribution, one is Poisson distribution and the other is exponential distribution.

The Poisson distribution is $f_i=e^{\lambda t}\frac{(\lambda t)^i}{i!}$, where $\lambda$ may be interpreted as the average number of change per unit time.

The exponential distribution is given by $g(x)=\mu e^{-\mu x}$, where $\mu$ is the rate-parameter or parameter of distribution.

By the given information $\lambda=\frac{2}{10}=\frac{1}{5}$ and $\mu=\frac{18}{60}=\frac{3}{10}.$

Now in the first $30$ minutes, number of customers arrive at the shop is $\lambda t=\frac{1}{5} \times 30=6$.

The value $\mu$ suggests that $\frac{3}{10}$ customer is served at $1$ minute.

So we have to find that no customers is served in $30$ minutes.

I am not sure how to do it.

Note: The correct answer is $\frac{2}{3}e^{-3}$.

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    $\begingroup$ Let $W$ the waiting time of the customer. In front of him he could have $0,1,2,...$ other customers whose waiting times $T$ are iid exponentials rvs. I would use the law of total probability and the fact that the sum of $n$ independent exponentials with the same parameter is a Gamma distribution to evaluate $P[W>30]=P[W=T_0,N=0]+P[W=T_0+T_1,N=1]+...$ where $N$ is the Poisson rv that count the number of customers in from of him. What do you think? $\endgroup$
    – Enrico
    Aug 27, 2023 at 8:22
  • $\begingroup$ @Enrico, What is $P[W=T_0, N=n]$ ? What are the random variables $T, N$ ? $\endgroup$
    – MAS
    Aug 27, 2023 at 11:22
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    $\begingroup$ $T_i$, $i=1,2,...$ are iid Exp($\mu$), N is Poisson($\lambda$). If $N=n$, then you have $P[W>30,N=n]=P[T_0+T_1+...+T_n>30,N=n]$. This is the generic term of the sum obtained with the law of total probability. Thus $P[W>30]=\sum_{n=0}^{\infty} P[T_0+...T_n>30|N=n]P[N=n]$. I forgot the ">30" in the previous comment. And $W:=T_0+...+T_n$ for a fixed $n$. Hope it makes sense my approach, even if I am not sure it is the correct one given the answer of Math1000. $\endgroup$
    – Enrico
    Aug 27, 2023 at 11:38

1 Answer 1

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The stationary distribution of this M/M/1 queue is $$ \pi_n = \frac13\left(\frac23\right)^n,\ n\geqslant0. $$ Conditioned on there being $n$ customers present, the waiting time of a customer has $\mathsf{Erlang}(n,\mu)$ distribution. So by the law of total probability, \begin{align} \mathbb P(W>30) &= \sum_{n=1}^\infty \mathbb P(W>30, N=n)\\ &= \sum_{n=1}^\infty \mathbb P(W>30\mid N=n)\mathbb P(N=n)\\ &= \sum_{n=1}^\infty \left(\int_{30}^\infty \frac{\frac3{10}\left(\frac 3{10}t\right)^{n-1}}{(n-1)!}e^{-\frac 3{10}t}\ \mathsf dt\right)\cdot\frac13\left(\frac23\right)^n. \end{align} Using Tonelli's theorem to interchange the sum with the integral, we have \begin{align} \mathbb P(W>30) &= \int_{30}^\infty \left(\frac1{15}e^{-\frac 3{10}t}\sum_{n=0}^\infty \frac{(t/5)^n}{n!}\ \mathsf dt\right)\\ &= \int_{30}^\infty \frac1{15} e^{-\frac t{10}}\ \mathsf dt\\ &= \frac 23 e^{-3}. \end{align}

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    $\begingroup$ I commented the OP with my approach that seems now wrong. Could you explain why using $\pi_n$ instead of using the Poisson distribution for $N$ is correct? Thanks. $\endgroup$
    – Enrico
    Aug 27, 2023 at 9:15
  • $\begingroup$ The method is new to me. Can you please tell me what probability is $\pi_n=\frac{1}{3} \cdot (\frac{2}{3})^n$ ? $\endgroup$
    – MAS
    Aug 27, 2023 at 11:27
  • $\begingroup$ $\pi_n=\lim_{t\to\infty}p_n(t),$ where $p_n(t)$ is the probability of there being $n$ customers in the system at time $t$. $\endgroup$
    – Math1000
    Aug 27, 2023 at 23:20
  • $\begingroup$ @Math1000 I am curious since I am studying these topics too: why using a Poisson, instead of the limit distribution, would be wrong? Thanks. $\endgroup$
    – Enrico
    Aug 29, 2023 at 7:45

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