If the probability that an individual suffers a bad reaction from an injection of a given serum is 0.001, then using Poisson's distribution, the probability that out of 2000 individuals, exactly 3 will suffer a bad reaction is:

Given:
​$$p = 0.001, \quad n = 2000 \\\lambda = np = 2000 \times 0.001 = 2 \\k = 3 \\$$​​
Formula used:
​$$P(X = k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!} \\$$​​
Solution:
​$$P(X = 3) = \frac{2^3 \cdot e^{-2}}{3!} = \frac{8 \cdot e^{-2}}{6} = \frac{4}{3} e^{-2} \\$$​​
Correct answer is (a)