A manufacturer of locks knows that 2 % of his | vedclass

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(A) Let $X$ be the number of defective locks in a box. Since the number of locks $n=100$ is large and the probability of a defect $p=0.02$ is small,we

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$1-5 e^{-2}$

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(A) Let $X$ be the number of defective locks in a box. Since the number of locks $n=100$ is large and the probability of a defect $p=0.02$ is small,we use the Poisson distribution with mean $\lambda = np = 100 	imes 0.02 = 2$. The probability of having $r$ defective locks is given by $P(X=r) = \frac{e^{-\lambda} \lambda^r}{r!} = \frac{e^{-2} 2^r}{r!}$. The manufacturer guarantees that there are not more than $2$ defective locks,meaning the box meets the quality if $X \le 2$. The box fails to meet the guarantee if $X > 2$. The probability of failure is $P(X > 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$. Substituting the values: $P(X > 2) = 1 - [\frac{e^{-2} 2^0}{0!} + \frac{e^{-2} 2^1}{1!} + \frac{e^{-2} 2^2}{2!}] = 1 - [e^{-2} + 2e^{-2} + 2e^{-2}] = 1 - 5e^{-2}$.

$A$ manufacturer of locks knows that $2 \%$ of his product is defective. If he sells the locks in boxes each with $100$ locks and guarantees that not more than $2$ locks will be defective in a box,then the probability that a box will fail to meet the guaranteed quality is

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