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.
Best Answer
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