I hear your cry for help. I think of many probability problems as a process of figuring out individual ways things happen and then accounting for each case. That's what we'll do here.
So, I will use the notation n choose k to denote the number of ways k objects can be selected from n, which is given by the formula n!/(k! (n-k)!). N! in turn is given by 1x2x3...xN. Hopefully this is all familiar to you.
So to determine the probability, we figure out how many ways our event could happen, then divide by the number of ways anything could happen. So, starting with
0) How many different ways are there to pick 5 marbles from a bag of 50?
There are 50 choose 5 = 50x49x48x47x46x45! / (5! x 45!) = 50x49x48x47x46 / 5! = 2118760 ways. By a strange not-at-all coincidence, if you type "50 choose 5" into google, this number comes up.
a) Probability of 2 red, 2 blue, and 1 of a third color.
20 choose 2 = 190
ways to pick the red,
10 choose 2 = 45
ways to pick the red, and 20 ways to pick any one of the 20 remaining balls. "And" in this case means multiply, so multiplying together, we find 190x45x20 = 171000 ways for this to occur.
Overall probability then, is 171000 / 2118760 = 0.0807 (approximately).
b) Probability of at least 2 red marbles
"At least" here is code for "solve it backwards". So how many ways are there to have fewer than 2 red marbles? Well, you could have zero or one.
To get zero, you have to pick all five from the 30 marbles that are not red. That can happen in
30 choose 5 = 142506 ways.
Or, you can pick one of the 20 red marbles, and then four from the 30 marbles that are not red. That can happen in
20 x (30 choose 4) = 20 x 27405 = 548100.
We interpret "or" as add, so there are
548100 + 142506 = 690606
ways to draw zero or one balls.
Ok, now we remember that we just figured out how many ways there are to draw AT MOST ONE red ball. How many ways are there to get two or more? Well, all of the rest of the ways you can draw five balls. So that would be
2118760 - 690606 = 1428154.
Overall probability of drawing at least 2 balls is given by
1428154 / 2118760 = 0.6741 (approx)
c) 2 marbles of one color, 2 of another, and 1 of a third.
Annoyingly, my best answer for this is to break into the case where you get 2 red marbles and the case where you don't. I'll let you do the calculating on this one, shall I?
How many ways are there to draw 2 red marbles, 2 blue marbles, and 1 of another color? Oh, right, we just figured that out for a). There are 171000. Nothing magic about blue, though, There are the same number of ways to get green or yellow. So we multiply our original number by three.
There are 3 x 171000 ways that could happen.
The other case is that there are not 2 reds. The only way that could happen is that we choose 2 each from 2 sets of 10, and then 1 from the remaining 30. There are
(10 choose 2) x (10 choose 2) x 30
ways to do that. However, it could be any two of the three colors, and there are 3 choose 2 ways to pick them, so we have to multiply by 3.
3 x 45 x 45 x 30
So the overall answer is that there are
3 x 171000 + 3 x 45 x 45 x 30
ways to draw 2 each of 2 colors.
And hopefully you know how to get the probability from there.
I hope that helps.