13. Given 10 points on a plane, no three of which are collinear, draw 4 line segments, each connecting two of the points. These line segments are chosen arbitrarily, and each has an equal chance of being selected. The probability that three of these line segments form a triangle with three of the given 10 points as vertices is $\frac{m}{n}$, where $m, n$ are coprime positive integers. Find $m+n$.
13. If 4 line segments among 10 points form a triangle, then there are $C_{10}^{3}$ ways to choose this triangle. For the 4th line segment, if one of its endpoints belongs to the above triangle, there are $3 \times 7=21$ ways to choose it; if both endpoints do not belong to the above triangle, there are $C_{7}^{2}=21$ ways to choose it.
Thus, the number of ways to choose 4 line segments that form a triangle is $C_{10}^{3}(21+21)=42 C_{10}^{3}$.
On the other hand, choosing 4 line segments from the possible $C_{10}^{2}=45$ line segments among 10 points has $C_{45}^{4}$ ways.
Therefore, $\frac{m}{n}=\frac{42 C_{10}^{3}}{C_{45}^{4}}=\frac{16}{473}$, hence $m+n=16+473=489$.