If the probability that an individual suffers a bad reaction from injection of a given serum is 0.001, determine the probability that out of 2000 individuals, more than 2 individuals will suffer from a bad reaction. (Use $e^{-2} = 0.1353$)

0.323

The correct answer is Option (2) → 0.323

Let X be the number of individuals suffering a bad reaction. Then X has a binomial distribution with $n = 2000, p=0.001$. However, the calculations would be very laborious using binomial distribution. Hence we can use Poisson approximation with $λ = np = (2000) (0.001) = 2$.

$P(X = r) =\frac{λ^re^{-λ}}{r!}=\frac{2^re^{-2}}{r!}$

$P(X > 2) = 1 - [P(0) + P(1) + P(2)] = 1 -\left[\frac{2^0e^{-2}}{0!}+\frac{2^1e^{-2}}{1!}+\frac{2^2e^{-2}}{2!}\right]$

$= 1-5e^{-2}=1-5(0.1353) = 0.323$