Probablity of drawing all the red balls while blue and green are still left Suppose that a box contains 10 red balls, 20 green balls, and 30 blue balls. 
Suppose also that balls are drawn from the box one at a time at random.
What is the probability that all the red balls are drawn before the blue or green balls are themselves exhausted. What is the probability that as the last red ball is drawn, there remains at least one blue and one green left in the box.
The answer I was given is $\dfrac{7}{12}$ and a general equation is:
$$
\dfrac{b g}{1-b}+\dfrac{b g}{1-g}
$$ where
$$
g=\dfrac{20}{60},b=\dfrac{30}{60}
$$
but why?
 A: Edit
The following gives the probability of draw the all red balls before any ball of other colour. It is not the asked answer and it is due to a misunderstanding. 
There is only one possibility to choose all $10$ red balls if we draw $10$ ball. The number of possible ways to draw $10$ balls is $\binom{60}{10}.$ Thus the probability is
$$\frac{1}{\binom{60}{10}}\approx 1.3263 \cdot 10^{-11}.$$
A: 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}$. 
A: You might be able to prove it by induction.  Let $P(R,B,G)$ be the probability that reds are the first to go given $R$ reds, $B$ blues and $G$ greens.  It is easy to check your formula for $P(1,1,1)$.  Note also that $P(R,B,0)=0$.
Now suppose your formula is true for $P(R,B,G)$ whenever $R+B+G\le N$.  Then $$P(R+1,B,G)=\frac{R+1}{R+1+B+G}P(R,B,G)+\frac{B}{R+1+B+G}P(R+1,B-1,G)+\frac{G}{R+1+B+G}P(R+1,B,G-1)$$
and now you can check the formula is true when $R+B+G=N+1$.
