jmc

We can use the idea of complementary probability to solve this problem without too much nasty casework. The probability that the committee has at least 1 boy and 1 girl is equal to 1 minus the probability that the committee is either all boys or all girls. The number of ways to choose a committee of all boys is $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$, the number of ways to choose a committee of all girls is $\left(\genfrac{}{}{0ex}{}{15}{5}\right)=3,003$, and the total number of committees is $\left(\genfrac{}{}{0ex}{}{25}{5}\right)=53,130$, so the probability of selecting a committee of all boys or all girls is $\frac{252+3003}{53,130}=\frac{31}{506}$. Thus the probability that the committee contains at least one boy and one girl is $1-\frac{31}{506}=\boxed{\frac{475}{506}}$.