For the case with two blue balls:
Define the probability of drawing a blue ball with one single draw as $p$ and the probability of the complementary event - no blue ball with a single draw as $1-p$.
You draw three times. Any distinguishable permutation with two blue balls and one ball with another colour (e.g. first draw blue, second draw other colour, third draw blue again) will have the same probability $P(B)$:
The number of distinguishable permutations is in general counted as:
where $n$ is the total number of elements in the permutation and $n_1, n_2...$ indicate the number of elements in subsets $1,2,....$ within which the elements cannot be distinguished from each other.
(that's the multinomial formula btw)
The way the problem presents itself here is that you have just two subsets, blue and non-blue. So you can apply the binomial formula which is a special case of the multinomial formula:
this is aka the Combination formula, and written as:
where $k$ is the number of elements in one of the subsets, here the one with blue balls.
Note that "being indistinguishable within the subset" is sometimes a bit difficult to grasp in the context of probability questions - as one could for example give both blue balls distinct numbers, e.g. 1 and 2, and with the formula above, for example the case with the first ball being blue ball 1 and the 2nd ball being blue ball 2, and the case with the opposite are counted together as one case only. However, this "summarized case" will get double the probability assigned.
Multiplying the number of distinguishable permutations with two blue balls with the probability of any such distinguishable permutation caclulated above gives the probability to get (exactly) two blue balls with three draws (event $X$):
(to generalize, you can write the probability that the number $X$ of blue balls drawn equals $k$ as follows:
which is the formula for the probability mass function of the binomial distribution (http://en.wikipedia.org/wiki/Binomial_distribution)).