jmc

Let $r$ represent the number of red balls in the bag. The probability that the first ball is red is $\frac{r}{10}$, while the probability that the other ball is red becomes $\frac{r-1}{9}$ (drawing two balls at random at the same time is similar to drawing one ball and then drawing another ball without replacement). So the probability that both balls are red is $\frac{r}{10}\cdot \frac{r-1}{9}$, which we set equal to $\frac{1}{15}$. $$\begin{array}{rl}\frac{r}{10}\cdot \frac{r-1}{9}& =\frac{1}{15}\Rightarrow \\ r(r-1)& =\frac{90}{15}\Rightarrow \\ r^2-r-6& =0\Rightarrow \\ (r-3)(r+2)& =0\end{array}$$ The value of $r$ cannot be negative, so $r=3$. There are $\boxed{3}$ red balls in the bag.