If the probability that an individual suffers a bad reaction from an injection of a given serum is 0.001. The probability that out of 2000 individuals, more than two individuals suffer from bad reaction is:

[Given that $e^{-2} = 0.13534$]

0.3233

The correct answer is Option (2) → 0.3233

Given: $n = 2000$, $p = 0.001$, probability of bad reaction $P(X>2) = ?$

Mean: $\lambda = n p = 2000 * 0.001 = 2$

Since $n$ is large and $p$ is small, use Poisson approximation:

$X \sim Poisson(\lambda = 2)$

$P(X>2) = 1 - P(X \le 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$

Poisson probabilities: $P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}$

$P(X=0) = \frac{2^0 e^{-2}}{0!} = e^{-2} = 0.13534$

$P(X=1) = \frac{2^1 e^{-2}}{1!} = 2 e^{-2} = 0.27068$

$P(X=2) = \frac{2^2 e^{-2}}{2!} = 2 e^{-2} = 0.27068$

$P(X \le 2) = 0.13534 + 0.27068 + 0.27068 = 0.6767$

$P(X>2) = 1 - 0.6767 = 0.3233$