A bag contains 3 red, 4 blue, and 5 green balls. Peter draws a ball from the bag, and then Angelina draws a ball. What is the probability that Angelina got a green ball?  
So far I have this:
Scenario A: 1st ball is not green, 2nd green: 7/12 * 5/11 = 35/132
Scenario B: 1st ball is green, 2nd green: 5/12 * 4/11 = 20/132
--> $$\frac{55}{132} = \frac{5}{12}$$
 A: Angela's draw is actually independent from the order she draws the ball whether she is first, second, ..., or twelfth (surprisingly enough). So, the probability of her drawing a green ball is actually $\dfrac{5}{12}$. To give you a better explanation, suppose there are $k$ green balls and $n$ total balls. Then, the probability that Angela draws a green ball is:
$$\require{cancel} \dfrac{k}{n}\dfrac{k-1}{n-1}+\dfrac{n-k}{n}\dfrac{k}{n-1} = \dfrac{k(k-1)+k(n-k)}{n(n-1)} = \dfrac{k \cancel{(n-1)}}{n \cancel{(n-1)}} = \dfrac{k}{n}$$
Which is the same probability that Peter has for drawing a green ball.
A: Your approach is fine, but it's worth realizing that unless you know something about the ball Peter has chosen, he may as well not have chosen at all, in which case Angelina simply has a $5/12$ chance of choosing a green ball.
Where this point really becomes worth understanding is if, for example, not only Peter, but also Olivia, Ned, Melissa, and Larry choose balls before Angelina.  The tree of possibilities is tedious and messy, but the answer is still $5/12$.
A: HINT
Brute force: condition on the color of the first drawn ball, there are 3 cases. Find the probability of each case happening, and what is the resulting probability in each case.
Then group them together in one final expression.
