A bag contains 10 marbles: 5 are green, 3 are red, and 2 are blue. Jose chooses a marble at random, and without putting it back, chooses another one at random. What is the probability that both marbles he chooses are red?
There are 3 red marbles out of the 10 total marbles, so the probability Jose chooses a red marble the first time is $\frac{3}{10}$. Then there will be only 9 marbles left in the bag. Assuming Jose's first choice was red, there will be 2 red marbles left among the 9 marbles. The probability Jose chooses a second red marble can therefore be expressed as $\frac{2}{9}$. To find the probability that both of these choices are red, we can multiply the probabilities: $\frac{3}{10} \times \frac{2}{9} = \boxed{\frac{1}{15}}$.
