Probability of drawing 2 marbles

[aaaeee123]

Homework Statement

There are m different buckets of marbles where each bucket has a different number of colored marbles (m). The total number of marbles is the same in each bucket and is K.

There are K individuals who walk past each bucket and pick 1 marble out of each one (so all marbles are given out).

What is the probability for an individual to have at least 2 colored marbles?

Homework Equations

The Attempt at a Solution

I have been trying to think of this as:

Prob(at least 2 colored marbles) = 1 - [prob(no colored marbles) + prob(exactly 1 colored marble)].

And where
prob(no colored marbles) = ((K-m[1])/K) * ((K-m[2])/K) * ... * ((K-m)/K)
prob(no colored marbles) = (1-(m[1]/K)) * ... * (1-(m/K))

I am having trouble figuring out the prob(exactly 1 colored marble). Is it from the binomial distribution with binomial parameter (sum_i(m))?

Also, from a larger standpoint, I am not sure this approach is correct due to the fact that the marbles are not replaced. The buckets are independent but clearly what the first person draws affects the second person, etc. So, should I be using a hypergeometric distribution?

Thanks in advance!

 

[JeSuisConf]

Well, by symmetry of the problem, no, it does not matter that there are k people (because the distribution of draws for any person is the same, so just imagine the first person.)

As for the probability of exactly 1 marble, it is a rather ugly beast. There are m ways it can happen, and the probability of each of those ways is different - and each way of it happening is calculated by the product of m terms. That is, it would look like

(m[1]/K)*(1-m[2]/k)*... * (1-m[n]/K) +
(1-m[1]/K)*(m[2]/k)*... * (1-m[n]/K) + ... etc.

 

[aaaeee123]

Thanks a lot. This makes a lot of sense (one of those feelings of man, why didn't I think of that). Basically it is the probability of getting exactly the first marble (and none of the others) plus the probability of getting exactly the second marble (and none of the others), plus the probability of getting exactly the third marble (and none of the others) etc, etc. Is that the right way to think about it?

Can you explain a bit more what you mean by the symmetry of the problem? I understand that the distribution of draws is the same for any person but it still seems to me that what the first person draws affects the other draws. If I come up with, for instance, the probability of the first person getting 2 or more colored marbles is 0.25. Can I say that I then expect 4 out of 10 people (on average) to get 2 or more colored marbles? If I wanted to test for this would a fisher exact test be inappropriate?

Thanks again for your reply!

Well, by symmetry of the problem, no, it does not matter that there are k people (because the distribution of draws for any person is the same, so just imagine the first person.)

As for the probability of exactly 1 marble, it is a rather ugly beast. There are m ways it can happen, and the probability of each of those ways is different - and each way of it happening is calculated by the product of m terms. That is, it would look like

(m[1]/K)*(1-m[2]/k)*... * (1-m[n]/K) +
(1-m[1]/K)*(m[2]/k)*... * (1-m[n]/K) + ... etc.