Customers at the bank

Question

Clients arrive in a bank according to a Poisson process of rate $\lambda>0$ for hour.

a) Knowing that after thirty minutes have arrived $3$ clients, find the distribution of number of clients arrived in the first $15$ minutes.

Let us assume that two clients have arrived in the first hour. Finding the probability that:

b) they are both arrived in the first $20$ minutes.

c) one of them at least has arrived in the first $20$ minutes.

For a) we have $X=($numbers of clients in the first $15$ minutes$)\sim Bin(3,\frac{1}{2})$.

For b) and c) I'm stuck. Could you please explain me how to approach the problem? Thanks in advance for any help!

Answer 1

Here are some ideas to think about that may help you:

For a person who is known to have arrived in the first hour, what is the probability that they arrived in the first $20$ minutes?

Also arrivals in a Poisson process are independent.

Answer 2

For a), let $T>0$ be a fixed real number, $n\geqslant 0$ a fixed integer, and $0<T_0<T$ a fixed real number. Then the distribution of $N(T_0)$ conditioned on $\{N(T)=n\}$ is found by "reversing" the conditioning: \begin{align} \mathbb P(N(T_0) = m\mid N(T)=n) &= \frac{\mathbb P(N(T_0)=m, N(T)=n)}{\mathbb P(N(T)=n}\\ &= \frac{\mathbb P(N(T)=n\mid N(T_0)=m)\mathbb P(N(T_0)=m)}{\mathbb P(N(T)=n)}\\ &= \frac{\mathbb P(N(T) - N(T_0) = n-m)\mathbb P(N(T_0)=m)}{\mathbb P(N(T)=n)}\\ &= \frac{e^{-\lambda(T-T_0)}\frac{(\lambda(T-T_0))^{n-m}}{(n-m)!} e^{-\lambda T_0}\frac{(\lambda T_0)^m}{m!}}{e^{-\lambda T} \frac{(\lambda T)^n}{n!}}\\ &= \frac{n!}{m!(n-m)!}\left(1-\frac{T_0}T \right)^n\left(\frac{T_0}T\right)^{n-m},\ 0\leqslant m\leqslant n, \end{align} i.e. $N(T_0)$ conditioned on $\{N(T)=n\}$ has a binomial distribution with parameters $n$ and $\frac{T_0}T$.

For b), let $n\geqslant 0$ be a fixed integer, $T>0$ a fixed real number, and $0<T_0<T$ a fixed real number. Then $$ \mathbb P(N(T_0) = n \mid N(T) = n) = \left(1-\frac{T_0}T\right)^n. $$

For c), the computation is similar: \begin{align} \mathbb P(N(T_0)>0\mid N(T) = n) &= 1 - \mathbb P(N(T_0)=0\mid N(T)=n)\\ &= 1 - \left(\frac{T_0}T\right)^n. \end{align}