
Combinatorics Question
I have two questions about probabilities when choosing things from a set.
Assume you have a bag of x red marbles, y blue marbles, and z green marbles.
1. If you choose 6 marbles from the bag, without replacement, what is the probability that you get exactly 2 of each type of marble?
2. If you choose 6 marbles from the bag, with replacement, what is the probability that you get exactly 2 of each type of marble?
I'm not really sure how to go about solving these.
Thanks for any help.

You must assume that each of x, y, & z is at least 2.
$\displaystyle { x \choose 2} = \frac {x!}{(x2)!(2!)}$ is the number of ways of choosing two reds marbles.
The total number of ways of choosing six marbles is $\displaystyle {x+y+z \choose 6}$ without replacement.
So, $\displaystyle \frac {{ x \choose 2}{ y \choose 2}{ z \choose 2}}{{x+y+z \choose 6}}$.
Now you try the next one.