# To compute probability in Poisson distribution?

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 $$\frac{2}{3}e^{-3}$$.

• 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? Aug 27, 2023 at 8:22
• @Enrico, What is $P[W=T_0, N=n]$ ? What are the random variables $T, N$ ?
– MAS
Aug 27, 2023 at 11:22
• $T_i$, $i=1,2,...$ 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. Aug 27, 2023 at 11:38

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}
• 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. Aug 27, 2023 at 9:15
• The method is new to me. Can you please tell me what probability is $\pi_n=\frac{1}{3} \cdot (\frac{2}{3})^n$ ?
• $\pi_n=\lim_{t\to\infty}p_n(t),$ where $p_n(t)$ is the probability of there being $n$ customers in the system at time $t$. Aug 27, 2023 at 23:20