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(B) Given $n = 2000$ and $p = 0.001$. Using the Poisson distribution, the parameter $\lambda = np = 2000 	imes 0.001 = 2$. The probability of $X$ ind

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

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(B) Given $n = 2000$ and $p = 0.001$. Using the Poisson distribution, the parameter $\lambda = np = 2000 	imes 0.001 = 2$. The probability of $X$ individuals suffering a reaction is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$. We need to find $P(X > 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$. Calculating the individual probabilities: $P(X=0) = \frac{e^{-2} 2^0}{0!} = e^{-2}$ $P(X=1) = \frac{e^{-2} 2^1}{1!} = 2e^{-2}$ $P(X=2) = \frac{e^{-2} 2^2}{2!} = 2e^{-2}$ Summing these: $P(X \le 2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5e^{-2}$. Therefore, $P(X > 2) = 1 - 5e^{-2} = 1 - \frac{5}{e^2}$.

$Y$	$-1$	$0$	$1$
$P(Y)$	$0.6$	$0.1$	$0.2$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\lambda$	$2\lambda$	$3\lambda$	$4\lambda$

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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