Simplifying a probability problem I am looking for a way on how to think about a probability problem. It feels like there's a way how to simplify the task.
So, here's example of the problem:

From a pool of $15$ balls ($1$ red, $2$ blue, $3$ green, $4$ yellow and $5$ black) a set of $5$ balls, at random, is selected and put in a bag. We don't know which $5$ balls are there.
Now, $2$ balls are at random picked from this set of $5$ balls, observed and put back with the other $3$.

Let's say we saw a blue and a yellow ball. Now i'd like to calculate probability of picking specific set of two balls (there are $14$ such sets).
One obvious approach I see is to list all the possible sets of $5$ balls what could be drawn from the initial pool of $15$ balls (there are $71$ such set), each with certain probability being drawn, $P(i)$; filter it to leave only those sets that contain at least one blue and one yellow ball (there are $25$ such sets), and then calculate the probabilities for each of the $14$ two-ball sets for each of those $25$ five-ball sets, weighing (multiplying) them by $P(i)$ and summing them up in $14$ sums.
The question I'd like to find out, is if there is simpler way to find the result avoiding so many iterations over specific sets.
Maybe there's some smart way how to think about those unseen $3$ balls in the bag, that would let me avoid iterating over the $25$ sets of $5$ balls that contain a blue and a yellow ball.
 A: Doing the problem assuming you saw a blue and yellow ball is easier than the general setting, in the sense that each case of color you see can be done with the same technique but the general pattern could be a bit ugly ; you would have to do them one by one (but of course after you understand the first one the other ones would be done similarly).
So you took $5$ balls out of $15$ and two of those $5$ balls are one blue and one yellow. You need to understand what's the probability distribution of the other three balls now. You also need to incorporate the info that you picked the two balls at random ; it is more likely to pick a blue and a yellow ball to look at if there is $2$ blue balls and $3$ yellow in the bag than otherwise, for instance. 
So say you want to compute the probability of picking a blue ball and a red ball in the second try. You need to look at all possible cases which make that happen with non-zero probability : there must be a red ball in the bag for that to happen, so the two cases would be $1$ blue ball inside or $2$. 
First case ; what is the probability that there is one red ball and one blue ball in the bag (precisely) assuming you picked one blue and one yellow after that?
$$
\mathbb P(1 R + 1 B \, | \, 1B + 1Y \text{ were picked } ) = \frac{\mathbb P( \{ 1R + 1B \} \cap \{1B + 1Y \text{ are picked } \})}{ \mathbb P(\{1B+1Y \} ) }.
$$
So the denominator is easy to compute ; the fact that you picked $5$ balls first then $2$ out of those $5$ is the same thing as just picking $2$ balls out of $15$ if we don't care about what's in the bag, but only about what two balls we pick ; so the probability to compute is the number of ways to pick $1$ blue ball out of two and $1$ yellow ball out of $4$. There are different ways to compute this. One of them is this : either you pick the blue first and the yellow second, or the opposite ; this accounts for all cases. You divide through by $2$ because you don't care about the order in which the balls appear, so you don't want to double count. So 
$$
\mathbb P( \{ 1B + 1Y \} ) = \frac{ \binom 21 \binom 41 + \binom 41 \binom 21 }{2\binom {15}2 } = \frac{16}{\binom {15}2} = \frac{32}{15 \cdot 14} = \frac{16}{105}.
$$
For the numerator, we should rewrite this as "$1R+1B+nY$ with $n \in \{1,2,3\}$" which gives us three cases to add up. Again, computing what happens at each step will give you probabilities. (At this point I'm tired of the numbers.)
You do the same for the $1R+2B$ case and when you both have them, in each case you compute the probability of pulling out a red ball and a blue ball assuming you know what's in the bag of $5$ balls ; add things up and you'll have your answer.
Hope that helps,
