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?
This question appears in Zhou's A Practical Guide To Quantitative Finance Interviews. It has already been asked here and here.
I have a question regarding the book's solution. Zhou defines $T_r$ as the round number when the last red ball is sampled, and similarly for $T_b$, $T_g$. The probability asked in the problem is then $P(T_r < T_b < T_g) + P(T_r < T_g < T_b)$. Next, $$P(T_r < T_b < T_g) = P(T_r < T_b \cap T_g = 60) = P(T_r < T_b | T_g = 60) P(T_g = 60).$$ $P(T_g = 60)$ is of course $30/60$, but I don't understand the derivation for $P(T_r < T_b | T_g = 60)$. Zhou writes:
Among the 30 red and blue candies, each candy is again equally likely to be the last candy, and there are 20 blue candies, so $P(T_r < T_b | T_g = 60) = \frac{20}{30}$.
I don't get why we can just forget about the 30 green candies. On the event $T_g = 60$, the last candy is green, but the times for the $29$ other green candies are still random. I'd like to see a more rigorous argument that $P(T_r < T_b | T_g = 60) = \frac{20}{30}$.