Urn balls without replacement, probability on nth position > An urn contains $w$ white and $b$ black balls. $n$ extractions without replacement are made. What are the probabilty of: * get a black ball on $i-th$ extraction? * get a black ball on $i-th$ extraction and white on $j-th$ extraction, with $j > i$? I know that the hypergeometric distribution doesn't care about position, so what is the math behind this problem?

Just imagine them all placed randomly in a row, their position won't change by extraction

Then P(a black ball is in **any** position) $=\frac{b}{w+b}$,

and P(a white ball is in **any** position) $= \frac{w}{w+b}$.

This directly gives the answer for the first part

* P(get a black ball on i−th extraction) $=\frac{b}{w+b}$

* For P(get a black ball on i−th extraction and white on j−th extraction), the logic is more subtle, the probabilities of a $B-W$ pair occupying any two positions will be the same, hence the same as $B-W$ occupying positions $1$ and $2$, $=\frac{bw}{(b+w)(b+w-1)}$