Can somebody please please help me with this probability problem.?

Well it's hard if you don't have marbles. But you have the next best thing, letters.

RRRR GGG BBBB

So let's pull some marbles out. If I get

RGB

that's no good. So I have to have at least two marbles the same color. But if I pull

RRR

that's no good either. So I have to have 2 of 1 color and 1 of the other. I don't like writing a lot so what I'll do if you can pick either one marble or the other is I'll write (R, G, B). That means "red or green or blue". So your options are

RR(G,B) <- remember this counts as two ways; "two reds and a green" or "two reds and a blue"

(G,B)RR

GG(B,R)

(B,R)GG

BB(R,G)

(R,G)BB

So that's 2 times 6, or 12 ways. You don't even need to look at the marbles in the bag because there's at least two of every color.

Possibilities are fractions. Just multiply them together.

1st marble:

RRRR GGG BBBB

3 greens out of 11 marbles = 3 / 11

2nd marble: You lost 1 green one when you picked it first.

RRRR GG BBBB

2 greens out of 10 marbles = 2 / 10 = 1 / 5

3rd marble: It doesn't matter. We already have our two green ones, and there's marbles left to draw. Any marble is okay here.

(3 / 11) (1 / 5) = 3 / 55

It's all fractions and luck