TITLE: Choosing Balls: $3$ Green, $5$ Blue and $4$ red
QUESTION [0 upvotes]: So this is a question someone asked me about (there's a nasty rumor that I know something about math) and that is ostensibly a basic combinatorics question, though I'm getting conflicting answers:
Say we have a bag with $3$ green, $5$ blue and $4$ red balls.  What is the probability of choosing three balls of the same color?
As near as I can tell, the best way to solve this is the following: 
How many ways can we choose $3$ of the same color? Well, we can choose all 3 green balls (this is $1$ way) we could choose $3$ blue balls ($5 \choose 3$ ways to do this), or we could choose $3$ red balls (4 ways). Dividing by $12\choose 3$ total ways to do this, we get something like .0681... as our probability.
On the other hand, we could say the probability is P(drawing 3 green)+P(drawing 3 blue)+P(drawing 3 red) which to me looks something like $$\frac{3\times 2\times 1}{12^3}+\frac{5\times 4\times 3}{12^3}+\frac{4\times 3\times 2}{12^3}\approx.0521...$$
So what am I missing here?
Thanks!

REPLY [1 votes]: You can also use the density function for the hypergeometric distribution for this.  http://en.wikipedia.org/wiki/Hypergeometric_distribution  Briefly, for two colors (say, white and black) dhyper(a,b,c,d) is the probability of drawing a white balls when there are b white balls in the urn, c black balls, and you are making d draws without replacement.
Your second approach is correct (once you fix the denominator) and boils down to:
dhyper(3,3,9,3)+dhyper(3,5,7,3)+dhyper(3,4,8,3)
As you are only making three draws, you don't have to worry about overlap.
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