Question

As a policy of quick serving restaurant. A customer complaint needs to be attended immediately. On an average on a given day 4 customers complaints. Find the following probabilities.

A) P ( No complaints)

B) P ( At most 2 complaints)

Answer 1

Poisson (X)=per 4 customer

the required probability $P(x)=\frac{e^{-\mu}\times \mu^n}{x!}$

x>1 so

$P(0)=\frac{e^{-4}\times 4^0}{1}=\frac{1}{e^4}$

Now, n=2

$P(x)=\frac{e^{-4}\times 4^2}{0!}=\frac{16}{e^{4}}$