Suppose that a book of 600 pages contains 40 | vedclass

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(C) The total number of pages is $600$ and the total number of mistakes is $40$. The average number of mistakes per page,denoted by $\lambda$,is $\lam

Vedclass · Accepted Answer

$e^{-2 / 3}$

Vedclass · Answer

(C) The total number of pages is $600$ and the total number of mistakes is $40$. The average number of mistakes per page,denoted by $\lambda$,is $\lambda = \frac{40}{600} = \frac{1}{15}$. The number of errors per page follows a Poisson distribution with parameter $\lambda = \frac{1}{15}$. The probability that a single page has no mistakes is given by $P(X=0) = \frac{e^{-\lambda} \lambda^0}{0!} = e^{-\lambda} = e^{-1/15}$. Since we are selecting $10$ pages at random,the probability that all $10$ pages have no mistakes is $(P(X=0))^{10} = (e^{-1/15})^{10} = e^{-10/15} = e^{-2/3}$.

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

Suppose that a book of $600$ pages contains $40$ print mistakes. Assume that these errors are randomly distributed throughout the book and the number of errors per page follows a Poisson distribution. The probability that all the $10$ pages selected at random have no print mistakes is

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