jmc

There are $\left(\genfrac{}{}{0ex}{}{9}{2}\right)=36$ pairs of points in the nonagon, and all but 9 (the sides of the nonagon) are diagonals, which means there are 27 diagonals. So there are $\left(\genfrac{}{}{0ex}{}{27}{2}\right)=351$ pairs of diagonals. Any four points on the nonagon uniquely determine a pair of intersecting diagonals. (If vertices $A,B,C,D$ are chosen, where $ABCD$ is a convex quadrilateral, the intersecting pair of diagonals are $AC$ and $BD$.) So the number of sets of intersecting diagonals is the number of combinations of 4 points, which is $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$. So the probability that a randomly chosen pair of diagonals intersect is $\frac{126}{351}=\boxed{\frac{14}{39}}$.