What is the probability of drawing 2 same colored marbles? Suppose there are 3 colored marbles (Red, Blue, Green). There are 10 red, 20 blue, 30 green marbles. You select 2 marbles without replacement. What is the probability that they are the same colors?
My solution:
$10 \choose 2$ ways to choose 2 red
$20 \choose 2$ ways to choose 2 blue
$30 \choose 2$ ways to choose 2 green
$60 \choose 2$ ways to choose either red, blue, or green marbles without replacement
So the probability is   $\frac{{10 \choose 2}+{20 \choose 2}+{30 \choose 2}}{60 \choose 2} $
I'm not sure if it makes sense to add the combinations in the numerator

There are two ways to think of this. The first is that if we define events $R$ to be two red balls, $B$ to be two blue balls, and $G$ to be two green balls, then we can verify that:


*

*$P(R) = \big(\frac{10}{60}\big)\big(\frac{9}{59}\big) = \frac{3}{118} \approx 0.025$

*$P(B) = \big(\frac{20}{60}\big)\big(\frac{19}{59}\big)= \frac{19}{177} \approx 0.107$

*$P(G) = \big(\frac{30}{60}\big)\big(\frac{29}{59}\big) = \frac{29}{118} \approx 0.246$


Then, $P(R \cup B \cup G) = P(R) + P(B) + P(G) = \frac{67}{177} \approx 0.379$
Note that each of the events $R,, B, G$ are disjoint and thus we may use the additive property. 
Second, if you would like to do this with counting then you simply compute $P(R)$ (and for the others) in terms of counting ratios as you did and note that they sum to precisely what you computed. You can cross check with what I presented here and you will find they are the same. 
