Drawing marbles from a bag without replacement. 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?
 A: 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.
A: 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.
