Candies withdrawal probability for a particular subsequence You are taking out candies one by one from a jar that has 10 red candies, 20 blue candies, and 30 green candies in it. What is the probability that there are at least 1 blue candy and 1 green candy left in the jar when you have taken out all the red candies?

I came across following solution but curious to see how others approach this problem:
Let $T_r$, $T_b$ and $T_g$ be the "draw number" of last red, blue and green candies drawn respectively. So the required probability here is $P(T_r<T_b \cap T_r< T_g)$
Now the event $T_r<T_b \cap T_r< T_g$ can be achieved in two different and independent ways, i.e., $T_r < T_b < T_g$ and $T_r < T_g < T_b$
$\therefore P(T_r<T_b \cap T_r< T_g) = P(T_r < T_b < T_g) + P(T_r < T_g < T_b) = \frac{20}{30} * \frac{30}{60} + \frac{30}{40} * \frac{20}{60} = \frac{7}{12}$
