Looking at the final balls is the way to go, as you can regard drawing all $60$ balls as choosing one of the equally probable permutation. Working from the back, you can also ignore a colour once it has been drawn.
The probability that the last ball is blue and that the last green comes after the last red is $\dfrac{30}{10+20+30}\times \dfrac{20}{10+20} =\dfrac{1}{3}$ or more generally $b \times \dfrac{g}{1-b}$.
The probability that the last ball is green and that the last blue comes after the last red is $\dfrac{20}{10+20+30}\times \dfrac{30}{10+30} =\dfrac{1}{4}$ or more generally $g \times \dfrac{b}{1-g}$.
So the probability that all reds are drawn before the final blue and final green are drawn is $\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}$ or more generally $ \dfrac{bg}{1-b} + \dfrac{bg}{1-g}$.
The question asks for combinations, but permutation seems more intuitive to me in this question, as the order of the marbles being drawn has to be taken into account.
I would use P(60,2) as the sample space. By doing so, I have to treat all 60 marbles as distinctive:
Blue: Marble 1, Marble 2, Marble 3..., Marble 20
Green: Marble 21, Marble 22, Marble 23..., Marble 40
Red: Marble 41 to Marble 60
(a) drawing two marbles of the same colour
$Probability = \frac{20P2+20P2+20P2}{60P2}$
(b) drawing a blue and a green marble
$Probability = \frac{(20P1*20P1)+(20P1*20P1)}{60P2}$
(c) drawing a blue or a green marble
$Probability = \frac{(20P1*40P1)+(20P1*40P1)}{60P2}$
Look forward to an answer that uses combinations instead.
Best Answer
Assumption: The first ball drawn is not put back in the bag.
Let E be the event that none of the two picked balls are blue. Further let $E_1$ and $E_2$ be the events that the first ball picked is not blue and the second ball picked is not blue respectively.
$P(E) = P(E_1 \cap E_2) = P(E_1)P(E_2 | E_1)$
$\implies P(E) = \left( {5 \choose 1}*1/7 \right) \left( {4 \choose 1}*1/6 \right)$
$\implies P(E) = 10/21 $