Chapter 9, Problem 9.1PTE

Customers arrive at a bank at a Poisson rate $\lambda$. Suppose that two customers arrived during the first hour. What is the probability that

a. both arrived during the first 20 minutes?

b. at least one arrived during the first 20 minutes?

To determine

To find: The probability that both the customers will arrive in first 20 minutes.

Answer to Problem 9.1PTE

The probability is 1/9.

Explanation of Solution

Given:

There are two customers that would arrive in first hour of bank.

The rate being followed is $\lambda$

Calculation:

It is given that two customers would come in first minutes. Thus,

  $20\text{ minutes = }\frac{1}{3}\text{hour}$

The probability that both the customers will arrive in first 20 minutes can be calculated as:

  $\begin{array}{c}P\left(N\left(\frac{1}{3}\right)=2|N\left(1\right)=2\right)=\frac{P\left(N\left(\frac{1}{3}\right)=2\cap N\left(1\right)=2\right)}{P\left(N\left(1\right)=2\right)}\\ =\frac{e^{-\lambda /3}{\left(\frac{\lambda }{3}\right)}^2}{2!}\times \frac{e^{-2\lambda /3}}{e^{-\lambda }\times {\lambda }^2}\times 2!\\ =\frac{1}{9}\end{array}$

Thus, the required probability is 1/9.

To determine

To find: The probability that at least one of the customers will arrive in first 20 minutes.

Answer to Problem 9.1PTE

The probability is 5/9.

Explanation of Solution

Calculation:

The probability that at least one of the customers will arrive in first 20 minutes can be calculated as:

  $\begin{array}{c}P\left(N\left(\frac{1}{3}\right)\ge 1|N\left(1\right)=2\right)=\frac{P\left(N\left(\frac{1}{3}\right)=0\cap N\left(1\right)=2\right)}{P\left(N\left(1\right)=2\right)}\\ =\frac{e^{-\lambda /3}{\left(\frac{\lambda 2}{3}\right)}^2}{2!}\times \frac{e^{-2\lambda /3}}{e^{-\lambda }\times {\lambda }^2}\times 2!\\ =\frac{5}{9}\end{array}$

Thus, the required probability is 5/9.