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The probability that an individual suffers a bad reaction from an injection is 0.001. The probability that out of 2000 individuals more than 2 suffer a bad reaction is

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$\frac{5}{e^{2}}$

$1 - \frac{5}{e^{2}}$

$1 - \frac{4}{e^{2}}$

$\frac{4}{e^{2}}$

answer is B.

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Detailed Solution

$λ = 2000 (0.001)$

$λ = 2$

$p (X > 2) = 1 - p (X = 0) - p (x = 1) - p (x = 2)$

We know ,from Poisson distribution :

$P (X = x) = \frac{λ^{x} e^{- λ}}{x!}$

$P (X = 0) = \frac{λ^{0} e^{- λ}}{0!} = \frac{1}{e^{2}} P (X = 1) = \frac{λ^{1} e^{- 2}}{1!} = \frac{λ}{e^{λ}} = \frac{2}{e^{2}} P (X = 2) = \frac{λ^{2} e^{- 2}}{2!} = \frac{λ^{2}}{2 e^{λ}} = \frac{2}{e^{2}}$

$p (X > 2) = 1 - \frac{5}{e^{2}}$

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