Probability, please help!?

Question

In how many ways can Kurt, Andrew, and John distribute 12 beers amongst the three of them? Assume it is possible for Kurt and Andrew to get 0 and John to get all twleve. But since Kurt is the biggest, Kurt will get at least 3 of the 12 beers.
Also, what is the probability Kurt will get at least 3 of the beers?

Answer 1

10/13 times Kurt will get three beers (unless of course you are counting your given, in which case, the question specifies that Kurt will ALWAYS get 3 beers).

There are total 429 possible combinations; 130 of those combinations have Kurt with atleast 3 beers.

Answer 2

Your question is inconsistent. It says that it is possible for Kurt (and Andrew) to get 0. But then it says that Kurt will always get 3. And it asks for the probability Kurt will get at least 3, which would be 100%... please clarify and I'll try to get you the answer.

I'll assume you meant *John* will always get at least 3 beers. So let's just give him 3 and then try and partition the remaining 9.

To partition the 9 items into 3 groups, you need 2 dividers. There are a total of 11 spots (9 beers and 2 dividers) to be filled. But order of the partitions doesn't matter, so it is a simple case of picking two spots to place the dividers:

11 choose 2 is 11 x 10 / 2 or 55 ways to partition the beers.

So that is your answer. There are 55 ways to divide the beers, if John always gets at least 3.

Now if you want to know the probability that Kurt will get at least 3 also, then imagine giving Kurt 3 (as well as John) and partitioning the remaining 6. This is 8 choose 2 or 8 x 7 / 2 or 28 ways.

So the chance that Kurt gets at least 3 beers is 28 out of 55. This is approximately 50.1%.