# Probability that 9 of 230 from 1 to 25,000 will be less than or equal to 22?

Suppose we choose randomly 230 different numbers from 1 to 25,000. What is the probability that at least 9 of them will be less than or equal to 22?

What is the probability that exactly 9 of them will be less than or equal to 22?

Thanks!

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If you draw with replacement, the probability that exactly 9 will be less than or equal to 22 is $\binom{230}{9}\left(\frac{22}{25000}\right)^9\left(\frac{24978}{25000}\right)^{221}$, where the binomial is the number of way to select which 9 are $\le 22$, the others are probabilities of $\le 22$ and $\gt 22$ or about $1.1\times 10^{-12}$ The extension to 0 through 8 should be easy to see.

If you draw without replacement, you have $\frac{\binom {22}{9}\binom {24978}{221}}{\binom {25000}{230}}=1.78*10^{-13}$

I think the difference is because drawing without replacement you use up lots of the ones below 22, so the probability is lower.

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using an hypergeometric distribution I got a different result (C(22,9) C(24978,221)) / C(25000,230) =~ 1.7*10^(-13). So which is the correct? –  anonymous Mar 19 '11 at 16:56
I believe both are correct, and it depends on the with or without replacement. You did specify different number in the original question, so it should be without replacement. –  Ross Millikan Mar 19 '11 at 17:07