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In some book, $250$ misprints are randomly and independently distributed on $500$ pages. What is the probability that there are no misprints on the first three pages?

I'm not sure how to solve this. Is using binomial distribution and adding the probabilities $P_1, P_2, P_3$ where $P_1$ is the probability that there is no misprint on the first page, $P_2$ on the second page and so on a good approach? How would you solve this problem? Any help would be much appreciated.

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It's not binomial. That we be correct if it were a case of sampling without replacement, but nothing says we can't have more than one misprint on a page. So a model of the problem is: we draw a number from $1$ to $500$ at random $250$ times. What is the probability that we never draw the number $1,\ 2\text{ or }3$?

Can you do it now?

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  • $\begingroup$ Is it $0.14$? I used the Poisson distribution $P(X = 0)=$ $\dfrac{2^0 \cdot e^{-2}}{1} = 0.14$ with $\lambda = 2$ and $x = 0$ (x being the number of errors we're interested in) $\endgroup$
    – Karla
    Commented May 20, 2020 at 7:41
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    $\begingroup$ I meant that the distribution is hypergeometric, so that the probability it $\left(\frac{497}{500}\right)^{250}\approx.22212$. If you want to use the Poisson approximation, then $\lambda=.5$, the average number of misprints per page, and the probability that there are no misprints on the first three pages is $\left(e^{-.5}\right)^3\approx.22313$ $\endgroup$
    – saulspatz
    Commented May 20, 2020 at 13:19

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