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A situation where there are n red balls and m white ones, which are taken out as described until some number r ≤ n of the balls have been removed. What is the probability that some number k ≤ n + m have been taken out, in terms of n, m, r and k? Briefly explain your rationale for your formula.

(Hint: A total of k balls will be withdrawn if there are r − 1 red balls in the first k − 1 withdrawals and the kth withdrawal is a red ball.)

This is a generalization of the previous problem: An urn holds 5 red balls and 3 white balls. They are drawn out one at a time (no replacement) until a total of 4 red balls have been taken out (and some unspecified number of white ones). Find the probability that exactly 6 balls have been taken out, showing the steps of your work.

I'm wondering how do you present the problem above and set it up.

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    Your 'some number $r\leq n$ of the balls have been removed.' should add 'red'? – 2017-01-27
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    I'm not entirely sure about this. Perhaps r is just a variable number less than the number of red balls (n). @Jan – 2017-01-27
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    No, I meant your general question says 'number $r\leq n$ of the balls have been removed', while the particular question says 'until a total of $4$ RED balls have been taken out' (my emphasis). Your current wording of the general question says until $r$ have been removed and asks what is the probability that $k$ have been taken out. The answer is trivial, $1$ if $r=k$ and $0$ if $r\neq k$. – 2017-01-27
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    @Jan... How would you formulate this into an answer for the given set of variables? – 2017-01-27

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This answer answers version of your question which reads (this is proper generalization of your earlier question):

There are $n$ red balls and $m$ white ones. Balls are drawn out one at a time (no replacement) until $r\leq n$ red balls have been removed. What is the probability that $k\leq n+m$ have been taken out, in terms of $n$, $m$, $r$ and $k$?

There are total $\binom{m+n}{m,n}$ ways to order the balls.

When $k^{\text{th}}$ ball is $r^{\text{th}}$ red one, $k\geq r$, $m-k+r\geq0$ and there are:

1) $m+n-k$ balls left, $n-r$ red and $m-k+r$ white,

2) $k-1$ balls have been drawn already, $r-1$ red and $k-r$ white.

The number of possible ways to order the balls in 1) and 2) is $\binom{m+n-k}{n-r,m-k+r}\binom{k-1}{r-1,k-r}$.