jmc

There's a $\frac{1}{3}$ chance that I will select each club. Let $n$ be the number of students in that club. There are $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ ways to choose a group of three students at a math club with $n$ members. Only $\left(\genfrac{}{}{0ex}{}{n-2}{1}\right)$ of these groups will contain the two co-presidents. Once I have selected that club, the probability that I give books to the co-presidents is $\frac{\left(\genfrac{}{}{0ex}{}{n-2}{1}\right)}{\left(\genfrac{}{}{0ex}{}{n}{3}\right)}$. Since the clubs have 5, 7, and 8 students, this means that the total probability is $$\frac{1}{3}\left(\frac{\left(\genfrac{}{}{0ex}{}{5-2}{1}\right)}{\left(\genfrac{}{}{0ex}{}{5}{3}\right)}+\frac{\left(\genfrac{}{}{0ex}{}{7-2}{1}\right)}{\left(\genfrac{}{}{0ex}{}{7}{3}\right)}+\frac{\left(\genfrac{}{}{0ex}{}{8-2}{1}\right)}{\left(\genfrac{}{}{0ex}{}{8}{3}\right)}\right)$$which after a bit of arithmetic simplifies to $\boxed{\frac{11}{60}}$.