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, exactly 3 individuals will suffer from a bad reaction. (Use $e^{-2} = 0.1353$)

0.180

The correct answer is Option (3) → 0.180

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 $λ = np = (2000) (0.001) = 2$.

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

$P(X=3)=\frac{2^3e^{-2}}{3!}=\frac{8}{6}e^{-2}=\frac{4}{3}(0.1353) = 0.180$