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?