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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. –  Jeff Fuller Dec 6 '12 at 3:46
    
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? –  Dilip Sarwate Dec 6 '12 at 3:50
    
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? –  Jeff Fuller Dec 6 '12 at 3:57
    
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. –  Dilip Sarwate Dec 6 '12 at 14:11
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2 Answers

There are $\displaystyle \binom {42}4$ ways of choosing $4$ marbles from $42$ without replacement. But to choose $2$ whites from $20$, $2$ blacks from $10$ and $0$ from the others you have $\displaystyle \binom {20}2 \displaystyle \binom {10}2\displaystyle \binom {6}0 \displaystyle \binom {6}0$. And the probability is $$\text P(\{\ 2 \ Whites, \ 2\ Blacks\ \})\cfrac {\displaystyle \binom {20}2 \displaystyle \binom {10}2\displaystyle \binom {6}0 \displaystyle \binom {6}0 }{\displaystyle \binom {42}4} $$ways

I think you can do the "one of each color" now.

This is a hypergeometric distribution, see this example.

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There are two approaches, which better yield the same answer. As Dilip Sarwate says, there are ${42 \choose 4}$ selections of marbles. How many ways are there to select two black marbles? How many ways to select two white marbles? Multiply them together and you have the number of ways to select two black and two white.

The other is in line with your comment. There are ${4 \choose 2}=6$ ways to order the two black and two white, and you already calculated the probability of pulling specifically white, white, black, black. Now multiply by six.

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does arrangement matter here, I think this is a hypergeometric distribution. –  user31280 Dec 6 '12 at 4:03
    
@F'OlaYinka: In the second case, the denominator is the number of ordered draws of 4 items from 42, so we have to count all the orders of the drawn marbles. Your calculation matches my first paragraph precisely and doesn't care about order because the denominator doesn't care. –  Ross Millikan Dec 6 '12 at 4:05
    
So basically, the probability of getting any arrangement of 2 B's and 2 W's is (20/42)(19/42)(10/42)(9/42), and then there are 6 specific arrangements, so you multiply by 6. I get it. I was concerned that the multiplication would change if we weren't looking at that specific "WWBB" permutation, but then I realized(duh) that multiplication commutes and the numerators/denominators are the same in every permutation. Thanks everyone! –  Jeff Fuller Dec 6 '12 at 4:07
    
$ $ $ $ $ $ $ $ gotcha!! –  user31280 Dec 6 '12 at 4:07
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