Q.6.12 - A First Course in Probability

Question

The number of people who enter a drugstore in a given hour is a Poisson random variable with parameter λ = 10. Compute the conditional probability that at most 3 men entered the drugstore, given that 10 women entered in that hour. What assumptions have you made? 

Step-by-Step Solution

Answer

The conditional probability that at most 3 men entered the drugstore is $0.2650$

Step 1: Content Introduction

The number of people who enter a drugstore in a given hour is a Poisson random variable with parameter λ = 10 .

Let X denote the number of people who enter a drugstore in given hour.

Step 2: Content Explanation

The probability density function of the Poisson distribution as shown below:

$P\left(x\right)=\frac{e^{-\lambda p}(\lambda p{)}^i}{i!}$

Let us assume that men and women are equally likely to come yo the store, which means $p=\frac{1}{2}$

Finding conditional probability that at most 3 men entered the drugstore given that 10 women in that one hour is:

$$\begin{gathered} P[X\le 3M\mid X=10W] \\ P[X\le 3M\mid X=10W]=\frac{P(\le 3M\cap 10W)}{P\left(10W\right)}, (X=10 \mathrm{W}) \\ P(X=10W)=\frac{e^{-\lambda p}{\left(\lambda p\right)}^x}{x!} \\ =e^{-5\frac{(5{)}^{10}}{10!}} \\ P(X\le 3M\cap X=10W) \\ P(X\le 3M\cap X=10W)=\left\{\begin{array}{l}P(0M,10W) + P(1M,10W) +\\ +P(3 M, 10 W)\end{array}\right.\left.\begin{array}{r}P(2M,10W)\\ \end{array}\right\} \\ =\frac{e^{-5}(5{)}^{10}}{10!}\times e^{-5}\left[\frac{(5{)}^0}{0!}+\frac{(5{)}^1}{1!}+\frac{(5{)}^2}{2!}+\frac{(5{)}^3}{3!}\right] \\ =e^{-10}\frac{(5{)}^{10}}{10!}[1+5+12.5+20.833] \\ =e^{-10}\frac{(5{)}^{10}}{10!}[39.3333] \\ P[X\le 3M\mid X=10W]=39.3333e^{-5} \\ =0.2650 \end{gathered}$$