Bernoulli Urn - Drawing white/black balls

Hi all,

I haven't done these kind of probability problems in a long time and trying to relearn. Here is a question that i'm having trouble with.

Q: An urn contains n white and m black balls which are randomly removed one at a time. If n > m, then what is the probability that there are always more white than black balls in the urn (until, of course, the urn is empty)? Explain why this probability is equal to the probability that the set of withdrawn balls always contains more white than black balls?

The answer is: (n-m)/(n+m)

I am thinking that the probability P(more white than black) is equivalent to the probability of drawing "(n-m) or less" white balls than black balls in a certain number of trials. now I am stuck. how do I proceed? Any help is appreciated.

Thanks

Re: Bernoulli Urn - Drawing white/black balls

Hey ronbeck.

Are you aware of the hyper-geometric distribution? You should be able to use this to help prove your result as it matches the process you are describing:

Hypergeometric distribution - Wikipedia, the free encyclopedia

Re: Bernoulli Urn - Drawing white/black balls

Hi Chiro,

Thanks for the suggestion. Turns out the problem I described is the classic Bertrand's ballot problem. It would def have taken me some time to come up with the solution.