I have a question, but I also have the solution. My problem is that I don't understand the solution!
Question: There are 10 red, 20 blue, 30 green balls in a bag. You keep removing balls at random. What is the probability that when you take the last red one out there is at least one green one and one blue ball remaining?
Answer: The easy way to do this is to reverse the order. The question is then what is the probability that a blue ball and a green ball are drawn before the first red ball. Once a ball of a given colour is drawn, it can be ignored as further draws don't affect the problem.
The solution then calculates the probability to this reverse order question to be 17/24.
How does this reverse order question relate to the original question?
Best Answer
Every ordered sequence of the sixty balls is equally likely. So every successful outcome for the original problem
$$\huge\color{red}{\bullet}\dots\color{blue}{\bullet}\color{red}{\bullet}\color{blue}{\bullet}\color{green}{\bullet}\, \color{green}{\bullet}$$
can be reversed to give a successful outcome for the new problem
$$\huge\color{green}{\bullet}\color{green}{\bullet}\color{blue}{\bullet}\color{red}{\bullet}\color{blue}{\bullet}\dots\color{red}{\bullet}$$