13) In an urn, there are 8 blue balls and 7 red balls. Mirco draws two balls, one after the other (without putting the first ball back into the urn before drawing the second). What is the probability that the two balls drawn are of the same color?
(A) $1 / 4$
(B) $1 / 2$
(C) $7 / 15$
(D) $8 / 15$
(E) none of the above
13. The answer is $\mathbf{( C )}$.

We need to calculate the probability $p_{1}$ that both balls drawn are blue and the probability $p_{2}$ that both balls drawn are red. The desired probability is the sum of $p_{1}$ and $p_{2}$. The probability $p_{1}$ can be calculated as follows: the probability that the first ball drawn is blue is $\frac{8}{15}$; if the first ball drawn is blue, at the time of the second draw, there are 7 blue balls and 7 red balls left in the bag, so the probability that the second ball drawn is blue is $\frac{7}{14}=\frac{1}{2}$. Therefore, we have $p_{1}=\frac{8}{15} \cdot \frac{1}{2}=\frac{4}{15}$.

Now let's calculate $p_{2}$ in a similar manner. The probability that the first ball drawn is red is $\frac{7}{15}$; if the first ball drawn is red, at the time of the second draw, there are 8 blue balls and 6 red balls left in the bag, so the probability that the second ball drawn is red is $\frac{6}{14}=\frac{3}{7}$. Therefore, we have $p_{2}=\frac{7}{15} \cdot \frac{3}{7}=\frac{1}{5}$.

We conclude by finding $p_{1}+p_{2}=\frac{4}{15}+\frac{1}{5}=\frac{7}{15}$, which is the desired answer.

[Problem proposed by S. Monica.]