jmc

The number of ways to draw out 3 balls from 15 is $\left(\genfrac{}{}{0ex}{}{15}{3}\right)=455$. We can choose 2 black balls and 1 white ball in $\left(\genfrac{}{}{0ex}{}{8}{2}\right)\left(\genfrac{}{}{0ex}{}{7}{1}\right)=196$ ways. We can pick 1 black ball and 2 white balls in $\left(\genfrac{}{}{0ex}{}{8}{1}\right)\left(\genfrac{}{}{0ex}{}{7}{2}\right)=168$ ways. Therefore we have $196+168=364$ ways to satisfy the condition, so the answer is $\frac{364}{455}=\boxed{\frac{4}{5}}$.