# Bernoulli Urn - Drawing white/black balls

• May 2nd 2013, 01:50 PM
ronbeck
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 Explain why this probability is equal to the probability that the set of withdrawn balls always contains more white than black balls?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
• May 2nd 2013, 05:26 PM
chiro
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
• May 2nd 2013, 07:10 PM
ronbeck
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.