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The probability that an individual suffers a bad reaction from an injection is 0.001. The probability that of 2000 individuals exactly three will suffers bad reaction is

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

$\frac{8}{3 e^{2}}$

$\frac{2}{3 e^{2}}$

$\frac{4}{3 e^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} λ = (0.001) 2000 = 2 \\ P (X = 3) = \frac{e^{- 2} (2)^{3}}{3!} \end{array}$

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