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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?


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up vote 1 down vote accepted

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? – the L 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

Sounds like this is probably a homework question, so my hint would be to model it as a binomial random variable with n=230.

What is the probability of success for each trial? What approximation can you make to the binomial distribution so that you dont have to take large factorials (i.e. 230!) ?

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thank you, it is not a homework problem, but an actual real life problem I ran into (and it has been a while since I took a probability course...) – the L Mar 19 '11 at 16:40
since we do not allow choosing the same number twice, how can I model this is a binomial variable? – the L Mar 19 '11 at 16:43
Actually, I reread your question. What I said only works if each trial can be any of the 25,000 numbers. This means you can have repeated numbers in your sample. The method will still give you an answer that is close because 25000 is so large, but if you need something closer, the hypergeometric distribution is probably what you are looking for. – Ralth Mar 19 '11 at 16:46

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