Suppose 5 red and 7 green balls are in a bag. Three balls are removed without replacement. Suppose $5$ red and $7$ green balls are in a bag. Three balls are removed without replacement. What is the probability that the second and third balls are both green?
I'm having trouble figuring out how to go about this problem. 
The way I see it is, that the first ball can be either red or green so P(R or G)?
Then the second and third ball must be green. So does this depend on the first choice. IS it a conditional probability?
Can I think of it this way:
$P(G)P(G)P(G) + P(R)P(G)P(G)$, i.e the probability of choosing green first then the other two being green plus the probability of choosing red first then the other two green?
 A: Since the colour of the first ball is not known, the probability of the second and third being green is the same as if you only took out two balls and want the probability of them both being green. In both cases, you are essentially choosing $2$ balls at random from the $12$ balls.
This simplifies the problem. The probability of the first green is $\dfrac{7}{12}$. The probability of the second green is then $\dfrac{6}{11}$. Multiplying gives the required probability:
$$\dfrac{7}{12} \dfrac{6}{11} = \dfrac{7}{22}.$$
A: Your intuition is correct. Here is a way to write it formally.
You wish to calculate $\mathbb{P}(X_2=G \cap X_3=G)$ which is equal to $\mathbb{P}(X_1 \in \{ R,G \} \cap X_2=G \cap X_3=G)$ (intersection with the whole space; $A = \Omega \cap A$).   
Now, $\mathbb{P}((X_1 = R \cup X_1 = G)) \cap X_2=G \cap X_3=G)$
$ = \mathbb{P}((X_1 = R \cap X_2=G \cap X_3=G) \cup (X_1 = G \cap X_2=G \cap X_3=G))$ (by DeMorgan's laws)
$= \mathbb{P}(X_1 = R \cap X_2=G \cap X_3=G) + \mathbb{P} (X_1 = G \cap X_2=G \cap X_3=G)$   (because the union is disjoint: you can't have $X_1=R$ and $X_1=G$ at the same time) 
Now, you have two probabilities to calculate and the easiest way to do it is by using the chain rule and conditional probabilities. For example, 
$\mathbb{P}(X_1 = R \cap X_2=G \cap X_3=G) = \mathbb{P}(X_1=R)*\mathbb{P}(X_2=G | X_1=R)*\mathbb{P}(X_3=G|X_2=G \cap X_1=R) = 5/12*7/11*6/10$
A: Either choose a single red followed by $2$ greens which is $5/12 * 7/11 * 6/10$ or choose $3$ greens which is $7/12 * 6/11 * 5/10$.  Sum those $2$ cases together and you get $7/22$. Interestingly, both cases are equiprobable at $7/44$ each.
Another method is to consider $12 \choose 3$ = $220$ possible ways to choose $3$ balls from $12$ and sum up the number of good ways which is ($1$ red followed by $2$ green) which is $5 \choose 1$ $7 \choose 2$ / $3$ = $35$ and then add that to how many ways to choose $3$ greens which is $7 \choose 3$ = $35$ so we get $70 / 220 = 7/22$.
The divide by $3$ in the first case is because we chose $1$ red and $2$ green but only RGG is a valid order but we could also get GRG and GGR so we counted $3$ times too many initially using just the binomials.
