Q:

Drawing 2 marbles from a box with 3 red, 3 purple, 5 green, and 7 blue marbles.

A box contains 3 red, 3 purple, 5 green, and 7 blue marbles. 2 marbles are selected from the box without replacement. What is the probability that you choose both marbles to be red or both marbles to be purple.
So far, I have figured out (I think):
$$\text{ Probability of both red  }= \frac{3}{\binom{18}{2}} = \frac{3}{(\frac{18!}{2!16!})} = \frac{3}{153}$$
Is this correct?

A:

You are correct. Another way to compute it:
Let $A_i$ be the event in which the $i$-th ball you grab is red. For the first ball you choose, you have $3$ red marbles out of a total of $18$. Therefore,
$$P(A_1) = \frac{3}{18}.$$ 
For the second marble you grab, given that you already took a red one, that is, given $A_1$, you have $2$ red balls out of a total of $17$, then
$$P(A_2 \mid A_1) = \frac{2}{17}.$$
Finally,
$$P(A_1 \cap A_2) = P(A_1)P(A_2 \mid A_1) = \frac{3}{18}\frac{2}{17} = \frac{1}{51}.$$
You can use the same reasoning to compute for the purple ones and then add those probabilities.

