Chapter B, Problem 1P

Summary Introduction

Interpretation: Probability of no customer arrival during a specific time and one and four customer arrival.

Concept Introduction:Probability of an event is how likely or how possibly an event take place. It shows an outcome of event which ranges between 0 and 1.

Answer to Problem 1P

The probability that during a specific minute no customer arrives is 0.1353. The probability that during a specific minute between one and four customer arrive is 0.8120

Explanation of Solution

Given information:

Number of customers arrive per minute = 2

In case of poisson distribution: The probability that $n$ customers will arrive in a system in time period $(T)$ , can be calculated using the following formula:

  $P_n=\frac{{(\lambda T)}^n}{n!}{e}^{-\lambda t}$

Where,

  $P_n$ = probability of n arrivals in T time periods

Probability of no customer arrival (n=0) during a specific time (T = 1)

  $\begin{array}{l}{P}_n=\frac{{(\lambda T)}^n}{n!}{e}^{-\lambda t}\\ P_n=\frac{{(2\times 1)}^0}{0!}{e}^{-2}\\ P_n=0.1353\end{array}$

The probability that during a specific minute no customer arrives is 0.1353

The probability that during a specific minute (T = 1) between 1 and 4 customer arrives using the following formula:

  $P_{0<n<5}=P_1+P_2+P_3+P_4$

  $P_{0<n<5}=\frac{{(\lambda T)}^1}{1!}{e}^{-\lambda t}+\frac{{(\lambda T)}^2}{2!}{e}^{-\lambda t}+\frac{{(\lambda T)}^3}{3!}{e}^{-\lambda t}+\frac{{(\lambda T)}^4}{4!}{e}^{-\lambda t}$

  $P_{0<n<5}=\frac{{(2\times 1)}^1}{1!}{e}^{-2}+\frac{{(2\times 1)}^2}{2!}{e}^{-2}+\frac{{(2\times 1)}^3}{3!}{e}^{-2}+\frac{{(2\times 1)}^4}{4!}{e}^{-2}$

  $P_{0<n<5}=0.8120$

The probability that during a specific minute between one and four customer arrive is 0.8120