60 balls in a bag 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?
 A: 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
$$\huge\color{green}{\bullet}\color{green}{\bullet}\color{blue}{\bullet}\color{red}{\bullet}\color{blue}{\bullet}\dots\color{red}{\bullet}$$
A: You can write out the letters corresponding to the colors of the balls taken out in order, and you get a 60-letter "word" consisting of Rs, Bs, and Gs. 
Notice that every 60-letter string is as likely to occur as its mirror image.
The probability that the last R has at least one B and one G to the right of it is therefore the exact same as the probability that the first R has at least one B and one G to its left. 
If you don't believe the first bolded statement, I recommend calculating the probabilities of some arrangements and their mirror images explicitly, but for a simpler version of the problem: 1 red ball, 2 blue balls, and 3 green balls. 
