in urn A white balls and B black balls. what would be the probability of taking the 5th ball being white the problem goes like that "in urn $A$ white balls, $B$ black balls. we take out without returning 5 balls. (we assume $A,B\gt4$) what would be the probability that at the 5th ball removal, there was a white ball while we know that at the 3rd was a black ball".
What I did is I build a conditional probability tree. as it seemed, it gets really ugly and parameters won't reduce, so the equation is huge, therefor probably it's not the correct path of solution.
I've got this intuition that the probability of the pth ball removal being black while qth ball removal being white is the same as the probability of the first ball being black, and the second ball being white - $\frac{A}{A+B-1}$ but this is only intuition and I can't explain it.
would appreciate your advising,
 A: The number of ways you could draw 5 balls with the 3rd being black is $B(A+B-1)_4$, where $(n)_r=n(n-1)\ldots (n-r+1)$. That's $B$ choices for the 3rd ball, and any length 4 sequence of the remaining $A+B-1$ balls. Similarly, the number of ways you could draw with the 3rd black and the 5th white is $AB(A+B-2)_3$. Divide them to get your formula.
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) $=\frac{B}{A+B}$,
and P(a white ball is in any position) $= \frac{A}{A+B}$.


*

*P(get a black ball on p−th extraction) $=\frac{B}{A+B}$

*For P(get a black ball on p−th extraction and white on q−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 $2$, $=\frac{AB}{(A+B)(A+B-1)}$ 

ADDED
If you are asking the conditional probability of the q-th ball being white, given that the p-th was black, $(q>p), $it will, of course, be $\dfrac{A}{A+B-1}$
