Consider a box with $n$ black and $m$ white balls, we randomly pick some $k$ of them(balls picked one by one), but if we pick white ball we get it back to box. Now we want to build a distribution for this task and find probability that we pick exactly $r$ black balls.
EDIT
My attempt : Suppose we pick some $i$ white balls , then there is $\binom{m+i-1}{i}$ (that's number of ways to pick it without order with returning) ways to pick it. Now there are $\binom{n}{k-i}$ (because we should choose $k-i$ balls from $n$ black balls) ways to pick black ball , so there are $\sum_{i=0}^{k}\binom{m+i-1}{i} \binom{n}{k-i}$ (all possible ways to pick them) ways to pick $k$ balls from box. Now the probability of choosing exactly $r$ black balls is : $\displaystyle \frac{\binom{n}{r}\binom{m+k-r-1}{k-r}}{\sum_{i=0}^{k}\binom{m+i-1}{i}\binom{n}{k-i}}$
I've edited some wrong assumption.
First of all : am I right ? If yes , does there some chances to simplify the sum ?
Let us assume that all balls are numbered and record the number and the color of the picked ball in each trial. Assume altogether $r$ black balls were picked up after $k$ trials. There are $\binom{n}{r}$ ways to choose the balls and $\binom{k}{r}$ ways to choose the trials which have given the balls. Besides the balls can be permuted between the chosen trials in $r!$ ways. The rest $k-r$ trials give white balls and this can happen in $m^{k-r}$ ways. Thus the overall number of $k$ trials resulting in $n$ black balls is $$ N(k,r)=r!\binom{n}{r}\binom{k}{r}m^{k-r}, $$ and the corresponding probability: $$ P(k,r)=\frac{N(k,r)}{\sum_{r=0}^kN(k,r)}. $$