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!