Probability of Marble Draws Formally 
Given a bag of six marbles with two colors, red and blue, what is the probability of drawing a set of three such that there is exactly one blue?
In the past, when given a problem such as this I would construct a tree diagram at each decision point and manually count the number satisfying the condition. But as examples become more complex, I find myself needing a more formal way of getting to the answer.
My thinking is that I have to solve the basic probability fraction of:
$\frac{\mathrm{num.\,of\,favorable\,outcomes}}{\mathrm{num.\,of\,total\,possible\,outcomes}}$
In our example, that would be:
$P(3\,marbles,\,1\,blue) = \frac{\mathrm{num.\,ways\,to\,draw\,3\,with\,one\,blue}}{\mathrm{num.\,ways\,to\,draw\,3\,marbles}}$
I think I can get the denominator with:
$_6{C}_3=20$
But for the numerator, absent of manually counting each possibility, I cannot think of a way.
 A: The story behind the actual numerator could go like this: How many ways of selecting a $n = 3 \text{-member}$ committee (read: drawing $3$ balls) from a group of $N = 6$ people (read: total number of $6$ balls) with $k= 1$ chairperson (read: $1$ blue ball) voted among $K =3$ candidates (read: $3$ total blue balls) to chair the committee? 
Well, given the $3$ candidates for chairmanship, there'll be ${K \choose k}={3 \choose 1}=3$ ways of selecting the chair. And for each of them, the $2$ ranking members among the $3$ non-chair-contenders can be chosen in ${N-K\choose n-k} = {3 \choose 2}$ ways. So the total number of ways is ${K \choose k}{N-K\choose n-k}=9.$

This experiment follows a hypergeometric distribution. From Wikipedia:

In probability theory and statistics, the hypergeometric distribution
  is a discrete probability distribution that describes the probability
  of $\displaystyle k$ successes in $\displaystyle n$ draws, without
  replacement, from a finite population of size $\displaystyle N$ that
  contains exactly $\displaystyle K$ successes, wherein each draw is
  either a success or a failure. 

So the probability of drawing $1$ blue among a set of $3$ is the 
$\small \Pr(1 \text{ success in } 3 \text{ draws without replacement}).$ 
In this case, 
$N=\text{total}=\text{blue and red}=6;$ 
$K =\text{total no. successes}= \text{blue}=3;$
$n=\text{draws}=3;$ and
$k =\text{successes in n draws}= 1.$
The denominator is as you calculated, ${N \choose n} ={ 6 \choose 3}=20$.
The numerator is $\large {K\choose k }{N-K\choose n-k}={3\choose 1 }{6-3\choose 3-1}={3\choose 1 }{3\choose 2}=3\times 3=9$
And the probability $\Pr(1 \text{ blue in } 3 \text{ draws})=9/20.$
In fact, your question is exactly equivalent to the urn experiment proposed as an example in Wikipedia.
