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.


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 ?

  • $\begingroup$ If you pick up the balls only once what is the sense in getting white balls back to box? May it be that you are interested in getting exactly $r$ black balls after say $N$ steps? $\endgroup$ – user Mar 7 '18 at 10:14
  • $\begingroup$ @user I pick up some $k$ balls. It make sense to return white balls. $\endgroup$ – openspace Mar 7 '18 at 10:30
  • $\begingroup$ May it be that you pick up the balls one by one and in the case if it is a white ball you put it back in the box, and $k$ is not the number of balls but the number of trials? $\endgroup$ – user Mar 7 '18 at 14:58
  • $\begingroup$ "Suppose we pick some $i$ white balls , then there is $A^i_m=m!/(m−i)!$ (because balls are get back to box) ways to pick it." This statement is wrong. There are $m^i$ such ways. Consider for example the case that all $i$ times the same white ball was picked up. $\endgroup$ – user Mar 8 '18 at 23:06
  • $\begingroup$ @user no , suppose we have three balls and we select two of them. Let $\{1,2,3\}$ be balls then we can make a pairs (1,2), (1,3) , (1,1) , (2,2) ,(2,3) and (3,3) , and that's equal to $3! /(3-2)! = 6$ $\endgroup$ – openspace Mar 9 '18 at 9:33

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: $$ P(k,r)=\frac{N(k,r)}{\sum_{r=0}^kN(k,r)}. $$

  • $\begingroup$ What is $l_{i}$? $\endgroup$ – openspace Mar 8 '18 at 14:54
  • $\begingroup$ And what is the problem in my answer? $\endgroup$ – openspace Mar 8 '18 at 15:17
  • $\begingroup$ @openspace I have added an explanation for the meaning of indices $l_i$. What concerns your answer I am not quite sure what do you mean with $A^i_m$. Have you already checked both versions? $\endgroup$ – user Mar 8 '18 at 15:51
  • $\begingroup$ $A_{m}^{i} = \frac{m!}{(m-i)!}$ $\endgroup$ – openspace Mar 8 '18 at 15:57
  • $\begingroup$ Please add in your question the idea behind using the value. I do not see how the possibility to draw the same white ball again is accounted for. $\endgroup$ – user Mar 8 '18 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.