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(D) Given: $n = 300$,$p = 0.01$. Let $m$ be the mean number of defective items. $m = n 	imes p = 300 	imes 0.01 = 3$. Since $n$ is large and $p$ is

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

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(D) Given: $n = 300$,$p = 0.01$. Let $m$ be the mean number of defective items. $m = n 	imes p = 300 	imes 0.01 = 3$. Since $n$ is large and $p$ is small,we use the Poisson distribution with parameter $m = 3$. The probability of $X$ defective items is given by $P(X = k) = \frac{e^{-m} m^k}{k!}$. We need to find the probability that the number of defective items is at most one,i.e.,$P(X \leq 1)$. $P(X \leq 1) = P(X = 0) + P(X = 1)$. $P(X = 0) = \frac{e^{-3} 3^0}{0!} = \frac{e^{-3} 	imes 1}{1} = \frac{1}{e^3}$. $P(X = 1) = \frac{e^{-3} 3^1}{1!} = \frac{3}{e^3}$. Therefore,$P(X \leq 1) = \frac{1}{e^3} + \frac{3}{e^3} = \frac{4}{e^3}$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

$A$ manufacturing company noticed that $1 \%$ of its products are defective. If a dealer orders for $300$ items from this company,then the probability that the number of defective items is at most one is

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