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 $e^{-2} \approx 0.13534$]

Given that $n = 2000$ individuals receive injections, and each has probability $p = 0.001$ of a bad reaction.

Since $n$ is large and $p$ is small, the Poisson approximation to the binomial distribution is appropriate.

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The parameter $\lambda$ for the Poisson distribution is:

$\lambda = n \times p$

$\lambda = 2000 \times 0.001$

$\lambda = 2$

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The probability that more than two individuals suffer from bad reaction is:

$P(X > 2) = 1 - P(X \leq 2)$

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

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Using the Poisson formula $P(X = k) = \frac{\lambda^k \times e^{-\lambda}}{k!}$ with $\lambda = 2$ and $e^{-2} = 0.13534$:

For $X = 0$:

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

$P(X = 0) = \frac{1 \times 0.13534}{1}$

$P(X = 0) = 0.13534$

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For $X = 1$:

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

$P(X = 1) = \frac{2 \times 0.13534}{1}$

$P(X = 1) = 0.27068$

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For $X = 2$:

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

$P(X = 2) = \frac{4 \times 0.13534}{2}$

$P(X = 2) = 0.27068$

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Therefore:

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$

$P(X \leq 2) = 0.13534 + 0.27068 + 0.27068$

$P(X \leq 2) = 0.6767$

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$P(X > 2) = 1 - P(X \leq 2)$

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

$P(X > 2) = 0.3233$