Question

Customers arrive at a certain retail establishment according to a Poisson process with rate λ per hour. Suppose that two customers arrive during the first hour. Find the probability that

(a) both arrived in the first 20 minutes;

(b) at least one arrived in the first 30 minutes.

Short answer

$$\begin{gathered} a)\frac{1}{9} \\ b)\frac{3}{4} \end{gathered}$$

Step-by-step solution

Part(a) Step 1: Given Information

We have to find the probability that both arrived in the first 20 minutes.

Part (a) Step 2: Simplify

Consider

$N\left(1\right)=2$.

$\begin{array}{rr}P\left(N\left(1/3\right)=2\mid N\left(1\right)=2\right)& =\frac{P\left(N\left(1/3\right)=2\right)P\left(N\left(2/3\right)=0\right)}{P\left(N\left(1\right)=2\right)}\\ & =\frac{\frac{{(\lambda /3)}^2}{2!}{e}^{-\lambda /3}\cdot \frac{{(2\lambda /3)}^0}{0!}{e}^{-2\lambda /3}}{\frac{{\left(\lambda \right)}^2}{2!}{e}^{-\lambda }}=\frac{1}{9}\end{array}$

Part (b) Step 1: Given Information

We have to find the probability that at least one arrived in the first 30 minutes.

Part (b) Step 2: Simplify

Calculating probability that no one arrived during the first half. The probability for that is

$\begin{array}{rr}P\left(N\left(1/2\right)=0\mid N\left(1\right)=2\right)& =\frac{P\left(N\left(1/2\right)=0\right)P\left(N\left(1/2\right)=2\right)}{P\left(N\left(1\right)=2\right)}\\ & =\frac{\frac{{(\lambda /2)}^2}{2!}{e}^{-\lambda /2}\cdot \frac{{(\lambda /2)}^0}{0!}{e}^{-\lambda /2}}{\frac{{\left(\lambda \right)}^2}{2!}{e}^{-\lambda }}=\frac{1}{4}\end{array}$

So, the probability that some of them arrived within the first half is localid="1648201989948" $\frac{3}{4}$.