Question

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

Answer 1

Answer:

The probability is 0.2650

Step-by-step explanation:

Let's start assuming that men and women come in at the same rate.

Let's define the following random variables :

X : ''Number of people that enter a drugstore''

M : ''Number of men that enter a drugstore''

W : ''Number of women that enter a drugstore''

The number of people will be the number of men plus the number of women

⇒

X = M + W

We are also assuming that M and W are independent random variables.

X ~ Po (10)

M ~ Po (λ1)

W ~ Po (λ2)

λ1 = λ2 because we assumed that men and women come in at the same rate.

λ1 = λ2 = λ

λ1 + λ2 = λ + λ ⇒ 2λ = 10 ⇒ λ = 5

M ~ Po (5)

W ~ Po (5)

Because X is the sum of two independent Poisson random variables.

We are looking for :

$P(M\leq 3/W=10)=P(M\leq 3)$

Because we assume independence.

$P(M\leq 3)= P(M=0)+P(M=1)+P(M=2)+P(M=3)$

$P(M=m)=\frac{5^{m}}{m!}e^{-5}$ because is a Poisson random variable with λ = 5

$P(M\leq 3)=\frac{5^{0}}{0!}e^{-5}+\frac{5^{1}}{1!}e^{-5}+\frac{5^{2}}{2!}e^{-5}}+\frac{5^{3}}{3!}e^{-5}$

$P(M\leq 3)=e^{-5}+5e^{-5} +(\frac{25}{2})e^{-5} +\frac{125}{6}e^{-5}=0.2650$

$P(M\leq 3)=0.2650$