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You have three bags, A, B, and C. Bags A and B hold x red and y blue marbles each; bag C holds 2x red marbles and 2y blue marbles.

Let us say that we pull out z marbles from bag A, z marbles from bag B, and 2z marbles from bag C. Will the expected number of red marbles pulled from bag A and bag B together be the same as the expected number of marbles pulled from bag C?

I believe the answer is yes, but I am having trouble figuring out why.

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2 Answers

up vote 2 down vote accepted

The expected number of red balls pulled from A is $z\frac{x}{x+y}$. This is because the probability that any of the $z$ individual balls is red is $\frac{x}{x+y}$. To see that, consider the experiment of taking $z$ balls. Number each ball you've taken - the first ball number $1$, the second ball $2$, and so on. Each of the numbers on its own is uniformly distributed, so the probability that the $k$th ball is red is exactly $\frac{x}{x+y}$.

The rest is an easy calculation, as in Ross's answer.

By the way, the number of red balls is distributed according to a Hypergeometric distribution (q.v.).

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Yes, but the probabilities are not independent. The probability that ball 2 is red depends on whether ball 1 was red, and so on. –  yrudoy Nov 11 '10 at 14:36
    
@yrudoy But all you want is the expectation, so it doesn't matter. Use linearity of expectation. –  Yuval Filmus Nov 11 '10 at 15:18
    
Hm. That would do it. Can you post a link to a proof of linearity of expectation? –  yrudoy Nov 11 '10 at 17:09
    
That's in every basic book (or lecture notes) about probability, and in the Wikipedia page about (mathematical) expectation. –  Yuval Filmus Nov 11 '10 at 22:38
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Yes. The expected number of red marbles from A is $\frac{xz}{x+y}$ and the same for B. The expected number from C is $\frac{4xz}{2x+2y}$ The 4 in the numerator is 2*2, one from the number of pulls, one from the number of red marbles.

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