A Hypergeometric Distribution Problem

[e(ho0n3]

Homework Statement

Balls are randomly withdrawn, one at a time without replacement, from an urn that initially has N white and M black balls.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 <= N, m <= M.

Homework Equations

A hypergeometric random variable with parameters n, N, m represents the number of white balls selected when n balls are randomly chosen from an urn that contains N balls, of which m are white. Its probability mass function is given by

p(i) = C(m, i) * C(N - m, n - i) / C(N, n)

The Attempt at a Solution

How can n white balls be withdrawn before m black balls? It could happen in the following ways:

- I draw out n balls and the all happen to be white,
- I draw out n + 1 balls and all but one is black,
- I draw out n + 2 balls and all but two are black, ...,
- I draw out n + m - 1 balls and all but m - 1 are black

The probability sought is the sum of the probabilities of each of the events described above right? If I draw k balls, n <= k < n + m, then the probability that n are white is C(N, n) * C(M, k - n) / C(N + M, k).

Here's the books answer, which I disagree with:

A total of n white balls will be withdrawn before a total of m black balls if and only if there are at least n white balls in the first n + m - 1 withdrawls. With X equal to the nuber of balls amount the first n + m - 1 withdrawn balls, then X is a hypergeometric random variable and

$$P\{X \ge n\} = \sum_{i=n}^{n+m-1} \frac{\binom{N}{i} \binom{M}{n+m-1-i}}{\binom{N+M}{n+m-1}}$$

I don't understand why the answer is P{X >= n} since this is the probability that at least n white balls are selected from n + m - 1 withdrawn balls.

 

[Nino MMP]

If at least n white balls are selected from the first n + m - 1 withdrawls, then it doesn't neccesarily mean all n white balls were withdrawn before m black balls were withdrawn. Could someone explain why the book answer is correct?