Question:
Balls are randomly withdrawn, one at a time without
replacement, from an urn that initially has
N white and M black balls. Find the probability
that n white balls are drawn before m black balls,
$n\leq N ,m \leq M$.
The answer:
Now I have a a lot of issues with this answer and I will try to summarize.
First:
I don't understand what the $m+n-1$ should represent.
Second:
The answer says $m$ or less; shouldn't it be just less than $m$? since if were to have at least $n$ balls then the largest number of black balls we are to have is simply $m-1$.
Third:
Overall as you've probably already noticed, I simply do not understand this solution whatsoever, for instance, why did we have to consider at least $n$ balls; shouldn't we just care about getting exactly $n$ balls before $m$?
Fourth:
Following up my third point I propose another solution where the experiment ends once we withdraw our $n^{th}$ ball before having withdrawn our $m^{th}$ ball; this can be done in a number of ways:
$n$ white; $0$ black
$n$ white; $1$ black; and so on until:
$n$ white; $m-1$ black
so the solution in this case should be:
$\sum\limits_{i=0}^{m-1} \frac{ \binom{N}{n} \binom{M}{i}}{\binom{M+N}{n+i}}$
Now obviously if you were to substitute with values for $n, m, N, M$; the results will be different.
What I want is to understand the author's answer first; then understand why my answer is incorrect.
Thanks in advance!
Best Answer
The correct answer is:
$$\sum_{i=n}^{m+n-1}{(\frac{\binom{N}{n-1} \binom{M}{i-n}}{\binom{M+N}{i-1}} \frac{N-(n-1)}{N+M-(i-1)})}$$
For explanation read the comments on the question.