6 red, 10 blue, and 2 green marbles Two marbles are selected without replacement. Determine the probability that 2 marbles of the same color are selected.  Do I just add the probabilities of 2 of each color being selected together?
 A: I’m assuming that the title gives the composition of the bag from which the marbles are to be selected. You could add the individual probabilities of selecting two red marbles, two blue marbles, and two green marbles, or you could calculate the desired probability all at once, like this.
There are $\binom62=15$ pairs of red marbles, $\binom{10}2=45$ pairs of blue marbles, and $1$ pair of green marbles, so there are $15+45+1=61$ pairs of identically-colored marbles. There are $18$ marbles altogether, so there are $\binom{18}2=153$ possible pairs. Thus, the desired probability is $\frac{61}{153}$.
Note, however, that the same calculation could be arranged in the form
$$\frac{\binom62}{\binom{18}2}+\frac{\binom{10}2}{\binom{18}2}+\frac{\binom22}{\binom{18}2}\;,$$
in which the first term is the probability of getting two red marbles, the second is the probability of getting two blue marbles, and the third is the probability of getting two green marbles, so you clearly do get the same result by adding the individual probabilities.
A: The direct probabilistic solution is to imagine the draws are made sequentially. So we want the probability of red then red or blue then blue or green then green. This is
$$\frac{6}{18}\cdot\frac{5}{17}+ \frac{10}{18}\cdot\frac{9}{17}+\frac{2}{18}\cdot\frac{1}{17}.$$   
