Chapter B, Problem 1P

Summary Introduction

Interpretation: The probability that during a particular time, no customer arrives. Further, customers arrive between one and four is to be determined.

Concept Introduction:

The probability between the arrival of one and four customers can be calculated by the Poisson distribution.

Explanation of Solution

The formula of the Poisson distribution is as follows,

  $\begin{array}{l}\text{Arrival rate (λ) = 2/minutes }\\ \text{Thus,Probability that a specific time, no customers arrives}\\ \text{By the formula,}\\ \text{P(0) = }\frac{{\text{λ}}^{\text{k}}{\text{e}}^{\text{-λ}}}{\text{K!}}\\ \text{where e is teh Euler's number}\\ \text{P(0)=}\frac{{\text{e}}^{-2}{\text{×2}}^{\text{0}}}{\text{1}}\\ \text{= 0}\text{.1353}\end{array}$

Probability of customer during one time period

  $\begin{array}{l}\text{Arrival rate (λ) = 2/minutes }\\ \text{Thus,Probability when 1 customer arrives }\\ \text{By the formula,}\\ \text{P(0) = }\frac{{\text{λ}}^{\text{k}}{\text{e}}^{\text{-λ}}}{\text{K!}}\\ \text{where e is teh Euler's number}\\ \text{P(0)=}\frac{{\text{e}}^{-2}{\text{×2}}^1}{\text{1!}}\\ \text{= 0}\text{.2706}\\ \end{array}$

The probability when 4 customer arrives

  $\begin{array}{l}\text{Arrival rate (λ) = 2/minutes }\\ \text{Thus,Probability when 4 customer arrives }\\ \text{By the formula,}\\ P(0)=\frac{{\lambda }^k{e}^{-\lambda }}{K!}\\ \text{where e is teh Euler's number}\\ \text{P(0)=}\frac{{\text{e}}^{-2}{\text{×4}}^1}{\text{4!}}\\ \text{= 0}\text{.0902}\\ \end{array}$

Probability when 1, 2, and 3 customers arrive at any specific time.

1= 0.2706

2=0.2706

3=0.1804

4=0.0902

Thus, the probability that between 1 and 4 customer arrives.

  $\begin{array}{l}\text{Sum of the probabilities}\\ 0.2706+0.2706+0.1804+0.0902\\ =\text{ 0}\text{.8118}\end{array}$

Thus,

The probability that no customer arrives is 0.1353

The probability that between 1 and 4 customer arrives is 0.8118