Question

A study of corruption for a certain geographic region showed an average of 5 corruptions occur per 20,000 people. In a city of 80,000 people, find the probability that at least 3 corruptions occur.

Answer 1

$P(X=x)= \frac{e^{-λ}λ^x}{x!} \\ λ= 5 \times \frac{80000}{20000}= 20 \\ P(X≥3) = 1- [P(X=0) + P(X=1) +P(X=2)] \\ = 1 -[\frac{e^{-20} \times 20^0}{0!} + \frac{e^{-20} \times 20^1}{1!} + \frac{e^{-20} \times 20^2}{2!}] \\ = 1 -e^{-20}(1 + 20 + 200) \\ = 1 -221 \times e^{-20} \\ = 0.99999$