Urn probability What is the easiest way to solve the following problem?
Given an urn with $B$ black balls and $R$ red balls, what is the probability that I pick the $r$'th black ball at the $k$'th trial if I am picking balls without replacement? 
Thanks
 A: Let $X$ represent a black ball and $Y$ represent a red ball, and suppose we perform $B+R$ trials. Then we can encode our experiment by writing down a string of $B+R$ letters involving $(B)$ $X$'s and $(R)$ $Y$'s that have been permuted in some fashion. Hence, the problem reduces to counting the number of strings where:


*

*The first $k-1$ balls contain $r-1$ black balls and $k-r$ red balls (that is, the first $k-1$ letters contain $(r-1)$ $X$'s and $(k-r)$ $Y$'s).

*The $k$th ball is black (that is, the $k$th letter is an $X$).

*The last $B+R-k$ balls contain $B-r$ black balls and $R-k+r$ red balls (that is, the last $B+R-k$ letters contain $(B-r)$ $X$'s and $(R-k+r)$ $Y$'s).


Putting everything together, we obtain:
$$
\dfrac{\binom{k-1}{r-1} \binom{B+R-k}{B-r}}{\binom{B+R}{B}}
$$
A: By Hypergeometric Distribution we know that 
$$\displaystyle P[X=k]=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$$
describes the probability to obtain $k$ successes in $n$ trials when there is a population $N$ and $K$ of them are of our interest.
First we consider the subproblem to obtain $r-1$ black balls in $k-1$ trials, then the probability of it is
$$\displaystyle P_s=\frac{\binom{B}{r-1}\binom{R}{k-r}}{\binom{B+R}{k-1}}$$
because we do not interest what is the order of these $r-1$ successes. But we interest that in the last trial, the $k^{th}$ trial, we have a success, so
$$\displaystyle P=P_s\frac{B-r+1}{B+R-k+1}=\frac{\binom{B}{r-1}\binom{R}{k-r}}{\binom{B+R}{k-1}}\frac{B-r+1}{B+R-k+1}$$
because before the $k^{th}$ trial we have $B+R-k+1$ balls in total, $B-r+1$ of them are black.
A: If the $r$th black ball is picked on the $k$th trial, then $r-1$ black balls have been picked in the preceeding $k-1$ trials.  No sequence of $r-1$ black balls and $k-r$ white balls is any more likely than any other, so you might as well compute the probability of picking all red on the first $k-r$ trials and black on the next $r$ trials, and then multiplying by $k-1\choose r-1$.  The probability of starting out with $k-r$ red balls is
$${R\over B+R}\cdot{R-1\over B+R-1}\cdots{R-(k-r-1)\over B+R-(k-r-1)}={R!/(R-(k-r))!\over(B+R)!/(B+R-(k-r))!}$$
and the probability of following this with $r$ black balls is
$${B\over B+R-(k-r)}\cdot{B-1\over B+R-(k-r)-1}\cdots{B-(r-1)\over B+R-(k-r)-(r-1)}={B!/(B-r)!\over(B+R-(k-r))!/(B+R-k)!}$$
Putting all this together gives the same answer as obtained already by Adriano.
