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Working on probability, just have a question that I can't get and I was looking for an explanation.

Suppose there is a box with 42 marbles. 20 white, 10 black, 6 red, 6 green. You choose 4 marbles at random without replacement. What is the chance you picked 2 white and 2 black? Also, what is the chance you get one of each color?

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    None of the techniques I already know seem to work. I would think it would be something like (20/42)(19/41)(10/40)(9/39), but that doesn't seem to make sense.2012-12-06
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    Hint: Sampling without replacement is the same as choosing a subset at random. There are $\binom{n}{k}$ different subsets of $k$ elements that can be chosen from a set of $n$ elements. So, how many different outcomes are there of the experiment of choosing four elements without replacement? How many of these consist of two white and two black elements?2012-12-06
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    So, there are $\binom{n}{4}$ total ways to get 4 marbles, and then there are $\frac{4!}{2 \cdot 2}$ favorable arrangements of B and W's?2012-12-06
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    While the answers posted give the complete solution, let me respond to your comment above. If the set of $4$ marbles can be subdivided into a set of $2$ white marbles and a set of $2$ black marbles, then the white marbles can be _any_ subset of size $2$ from a set of $k_1$ white marbles while the black marbles can be _any_ subset of size $2$ from a set of $k_2$ marbles. So the probability is $$\frac{\binom{k_1}{2}\binom{k_2}{2}}{\binom{n}{4}}.$$ There are _no_ favorable **arrangements.** You are dealing with sets, not marbles arranged in rows where you can talk of first marble etc.2012-12-06

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