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%.
