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