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An urn contains $w$ white and $b$ black balls. $n$ extractions without replacement are made (Hypergeometric distribution). The distribution of $\mathbb{P}(X_i=s)$ with $i\geq s$ ($s$ white on $ith$ drawn) is:

$$\mathbb{P}(X_i=s)=\frac{\dbinom{w}{s}\dbinom{b}{i-s}}{\dbinom{w+b}{i}}$$

How can I find $\mathbb{P}(Z=X_i+X_j)$? I know that $X_i$ and $X_j$ are not independent.

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  • $\begingroup$ "$n$ extractions without replacement are made" - that by itself does not imply a hypergeometric distribution. If you say then that the $Y$ is the number of black balls for example, then you can say $Y$ follows a HG distribution. $\endgroup$
    – Em.
    Jun 19 '16 at 10:32
  • $\begingroup$ @probablyme: I think here $X_i$ is the total number of white balls after $i$ draws and so has a hypergeometric distribution $\endgroup$
    – Henry
    Jun 19 '16 at 10:46
  • $\begingroup$ @Henry Oh, sorry for the confusion. I was making a point about semantics. Like often you will see people say that they roll a die and that this follows a uniform distribution. No, rolling a die does not follow anything. The value that shows on a fair die when you roll it follows a uniform distribution. $\endgroup$
    – Em.
    Jun 19 '16 at 10:50
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I suspect that if $i \le j$ and $0 \le z \le i+j$ with $Z=X_i+X_j$ then $$\mathbb{P}(Z=z) = \frac{\displaystyle \sum_{s: \max(0,z-w) \le s \le \min(i,z/2)} \dbinom{w}{s}\dbinom{b}{i-s}\dbinom{w-s}{z-2s}\dbinom{b-i+s}{j-i-z+2s}}{\dbinom{w+b}{i} \dbinom{w+b-i}{j-i}} $$ and I would guess that it might be difficult to simplify this except in special cases

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  • $\begingroup$ Hi @Henry, I'll think about it, thanks for answer. $\endgroup$
    – Paul
    Jun 19 '16 at 14:33

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