The problem is very poorly defined. It is a very small wonder that you are struggling with it.
Are we making three (3) draws? It doesn't say so. It should. I'll assume that is the case.
In that order? Will Blue,Red,White count? It doesn't say so. It should. I'll assume that is the case.
Do we put back the gumballs we pick or do we just go on? It doesn't say so. It should. I'll assume that we don't put things back between draws.
"equal amounts" is also very much insufficient.
If "equal amounts" means zero (0) each, then Pr(Red,White,Blue) = 0.
If "equal amounts" means one (1) each, then Pr(Red,White,Blue) = 1.
If "equal amounts" means some relative small number, then Pr(White|Red) can be significantly different from Pr(White). Let's try 2. Pr(Red,White,Blue) = 0.40. I won't bother to tell you how I found that.
If "equal amounts" means some really large number, then we have a TRINOMIAL DISTRIBUTION for which all probabilities can be calculated easily. Expand (R + W + B)<sup>3</sup> to get:
R<sup>3</sup>+3*R<sup>2</sup>*W+3*R<sup>2</sup>*B+3*R*W<sup>2</sup>+6*R*W*B+3*R*B<sup>2</sup>+W<sup>3</sup>+3*W<sup>2</sup>*B+3*W*B<sup>2</sup>+B<sup>3</sup>
R = Pr(Red on any draw) = 1/3
W = Pr(White on any draw) = 1/3
B = Pr(Blue on any draw) = 1/3
Picking out the term with all three: 6*R*W*B ==> 6*(1/3)<sup>3</sup> = 6/27 = 2/9 = 0.222222...
How close were you? How close am I?
With a problem statement like this, I'm afraid I would have to prevail upon the teacher to accept any reasonable set of assumption and an answer consistent with those assumptions.