*35*

### Exercise

The number of customers served by a bank’s central counter is 2 customers per minute. What is the probability that the counter will:

1. Serve 5 customers in a minute?

2. Serve 4 customers in the next 3 minutes?

### Solution

This problem involves a **Poisson distribution**, as we are dealing with the number of events (customers served) within a fixed period of time.

The Poisson probability mass function is given by:

\[

P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}

\]

Where:

– \( X \) is the number of customers served,

– \( \lambda \) is the average number of customers served in the given time period,

– \( k \) is the number of customers we are interested in,

– \( e \) is the base of the natural logarithm, approximately 2.71828.

In this problem:

– For Question 1, \( \lambda = 2 \) (as 2 customers are served per minute),

– For Question 2, \( \lambda = 6 \) (as 2 customers per minute over 3 minutes gives \( 2 \times 3 = 6 \)).

#### Question 1: Probability of serving 5 customers in one minute.

We are asked to find \( P(X = 5) \) for \( \lambda = 2 \):

\[

P(X = 5) = \frac{2^5 e^{-2}}{5!} = \frac{32 \cdot e^{-2}}{120} = \frac{32}{120 \cdot e^2}

\]

Calculating this:

\[

P(X = 5) \approx \frac{32}{120 \times 7.3891} \approx \frac{32}{886.692} \approx 0.036

\]

So, the probability of serving 5 customers in a minute is approximately **0.036**, or **3.6%**.

#### Question 2: Probability of serving 4 customers in the next 3 minutes.

We are asked to find \( P(X = 4) \) for \( \lambda = 6 \):

\[

P(X = 4) = \frac{6^4 e^{-6}}{4!} = \frac{1296 \cdot e^{-6}}{24}

\]

Calculating this:

\[

P(X = 4) \approx \frac{1296}{24 \times 403.4288} = \frac{1296}{9682.2912} \approx 0.134

\]

So, the probability of serving 4 customers in the next 3 minutes is approximately **0.134**, or **13.4%**.

—

### Summary of Answers

1. The probability of serving 5 customers in a minute is approximately **3.6%**.

2. The probability of serving 4 customers in the next 3 minutes is approximately **13.4%**.