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If the probability that an individual will suffer a reaction from an injection of a drug is 0.001 , then the probability that out of 2000 individuals having that injection, more than 2 individuals will suffer a reaction, is
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The correct answer is:
$1-\frac{5}{e^2}$
Here, $n=2000, P=0.001, \lambda=n p=2$
$\begin{aligned}
P(X>2) & =1-[P(X=0)+P(X=1)+P(X=2)] \\
& =1-\left[\frac{2^0 e^{-2}}{0 !}+\frac{2^1 e^{-2}}{1 !}+\frac{2^2 e^{-2}}{2 !}\right] \\
& =1-\left[e^{-2}+2 e^{-2}+2 e^{-2}\right]=1-5 e^{-2}=1-\frac{5}{e^2}
\end{aligned}$
$\begin{aligned}
P(X>2) & =1-[P(X=0)+P(X=1)+P(X=2)] \\
& =1-\left[\frac{2^0 e^{-2}}{0 !}+\frac{2^1 e^{-2}}{1 !}+\frac{2^2 e^{-2}}{2 !}\right] \\
& =1-\left[e^{-2}+2 e^{-2}+2 e^{-2}\right]=1-5 e^{-2}=1-\frac{5}{e^2}
\end{aligned}$
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