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$$P(X=i)=\frac{\binom{m}i\binom{N-m}{n-i}}{\binom{N}{n}}, i=0,1,\ldots, \min(n,m).$$

The above probability represents suppose that a sample of size $n$ is to be chosen randomly (without replacement) from an urn containing $N$ balls, of which $m$ are white and $N-m$ are black. If we let $X$ denote the number of white balls selected, then the probability of getting exactly $i$ white balls is $P(X=i)=\frac{\binom{m}i\binom{N-m}{n-i}}{\binom{N}{n}}, i=0,1,\ldots, \min(n,m).$

My question is:

What does the following probability represent $$\big(\frac{\binom{m}i\binom{N-m}{n-i}}{\binom{N}{n}}\big)^2$$?

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    $\begingroup$ Well, "doing it twice"? That is, you require that you draw your sample with exactly $i$ White and then you throw all the balls back in the urn and you do it again. $\endgroup$ – lulu Jan 31 '17 at 12:20
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    $\begingroup$ Yes as lulu is saying doing the same from same initial conditions so you need to back the balls up. $\endgroup$ – mathreadler Jan 31 '17 at 12:30
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That would be the probability of getting exactly $i$ white balls in each of two (independent) draws of $m$ balls, meaning that you replace the balls you drew in between the two draws.

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    $\begingroup$ it's the same. Suppose you have two identical urns and two people doing the same experiment. Then the probability of both getting exactly i white balls is as ABC asked $\endgroup$ – Julio Maldonado Henríquez Jan 31 '17 at 12:40
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    $\begingroup$ two experiments, so $X_1,X_2$, then the probability searched is $P(X_1=i,X_2=i)$ $\endgroup$ – Julio Maldonado Henríquez Jan 31 '17 at 13:03
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    $\begingroup$ @JulioMaldonadoHenríquez For a small trial, I found that $$\big(\frac{\binom{m}i\binom{N-m}{n-i}}{\binom{N}{n}}\big)^2\ne \frac{\binom{2m}{2i}\binom{2N-2m}{2n-2i}}{\binom{2N}{2n}}.$$ Why "combining all the balls from the two experiment and calculating the probability of getting exactly $2i$ white balls" is not same as of "having two identical urns and two people doing the same experiment, then calculating the probability of both getting exactly $i$ white balls"? $\endgroup$ – ABC Jan 31 '17 at 13:12
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    $\begingroup$ 'cause when you combine balls lose the independence. That's why you need two urns. $\endgroup$ – Julio Maldonado Henríquez Jan 31 '17 at 13:18
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    $\begingroup$ yes, just specify that $\lbrace X_j\rbrace_{j=1}^{10}$ are independient experiments equal to $X$ $\endgroup$ – Julio Maldonado Henríquez Jan 31 '17 at 13:46

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