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

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

Class 12 Mathematics Probability Probability distribution

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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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$e^{-0.3}$
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$e^{-0.5}$
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$e^{-1}$
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$e^{-2}$

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Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(e)$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$

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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