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Suppose there are $m$ red balls and $n$ blue balls in an urn. We randomly choose $p:m<p<n$ balls uniformly from the urn. What is the probability that exactly $q$ red balls are chosen?

Note:- Normally the answer would be $\frac{{m}\choose{q}}{{m+n}\choose{q}}$. However, since the number of balls that are chosen are provided, it is confusing me out. Any hints for the answer will also be appreciated.

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2 Answers 2

What you describe is a Hypergeometric Distribution of sampling without replacement.

It follows that

$$P(X=q)=\frac{{{m}\choose{q}}{{n}\choose{p-q}}}{{{m+n}\choose{p}}}$$

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This is true for a given $p$, but $p$ ranges from $m+1$ to $n-1$, so there is still work to do. –  Ross Millikan Jun 17 '13 at 2:23
    
@Ross Millikan I don't think the range of the variable p matters here. –  Sourav Jun 17 '13 at 2:25
    
I think it does. Suppose $q=1$. Then (if $n,m$ aren't too different) the probability of success will be higher if $p$ is at the small end of the range. So I think you need to use the fact that the distribution is uniform over the range, which I don't know how to do. –  Ross Millikan Jun 17 '13 at 2:40
    
@RossMillikan I interpreted the question to mean that there is a given $p:m\lt p\lt n$ but I see now that that would be an assumption but it must be true for any given sampling. –  Dale M Jun 17 '13 at 4:56
    
I am coming to your reading. In any case your expression is correct for a given $p$. –  Ross Millikan Jun 17 '13 at 13:06

This is not true. Take $q=1,m=2,n=6$ for example. For $p=3,5$ the probability of one red ball is $\frac {{2 \choose 1}{6 \choose 2}}{8 \choose 3}=\frac {30}{56}$. For $p=4$ it is $\frac {{2 \choose 1}{6 \choose 3}}{8 \choose 4}=\frac {40}{70}.$ The average of these if $\frac {23}{42}$, but your expression yields $\frac 14$. Dale M gives the correct expression for a given $p$.

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