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(C) The average number of errors per page is $\lambda_{page} = \frac{250}{500} = 0.5$. For a sample of $n = 2$ pages,the average number of errors is $

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$e^{-1}$

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(C) The average number of errors per page is $\lambda_{page} = \frac{250}{500} = 0.5$. For a sample of $n = 2$ pages,the average number of errors is $\lambda = n 	imes \lambda_{page} = 2 	imes 0.5 = 1$. According to the Poisson distribution,the probability of finding $X$ errors is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$. For no errors,we set $k = 0$: $P(X=0) = \frac{e^{-1} 	imes 1^0}{0!} = e^{-1} 	imes 1 = e^{-1}$.

In a book of $500$ pages,it is found that there are $250$ typing errors. Assume that Poisson law holds for the number of errors per page. Then,the probability that a random sample of $2$ pages will contain no error,is :

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