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?
 A: 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 $2$, $=\frac{bw}{(b+w)(b+w-1)}$    
A: *

*Number of all $i$ extractions are $S(i)=\binom{n}{i}*i!$

*Number of cases where to get a black ball on $i$th extraction is $B(i)$
$B(i)=b*\binom{n-1}{i-1}(i-1)!$
probability is $P(i)=\frac{B(i)}{S(i)}$


*

*Number of cases where to get a black ball on $i$th then white on $j$th is $W(i)$


$W(j)=b*w*\binom{n-2}{j-2}(j-2)!$
probability is $P(j)=\frac{W(j)}{S(j)}$
A: Simplest way to visualize this (at least for me) is to forget the fact that there are balls of two colors in the urn, and just imagine that each ball in the urn is a 'mixture' of both the colors.
Now, it is easy to see that each draw doesn't change any thing w.r.t. the distribution. That way the probability is independeny of each draw.
