7. From the sides and diagonals of a regular 12-sided polygon, three different segments are randomly selected. The probability that their lengths are the side lengths of a triangle is $\qquad$
7. $\frac{223}{286}$.

Let the diameter of the circumscribed circle of a regular 12-sided polygon be 1. Then, the lengths of all $\mathrm{C}_{12}^{2}=66$ line segments fall into 6 categories, which are
$$
\begin{array}{l}
d_{1}=\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}, \\
d_{2}=\sin 30^{\circ}=\frac{1}{2}, \\
d_{3}=\sin 45^{\circ}=\frac{\sqrt{2}}{2} \text { (please note) } \\
d_{4}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}, \\
d_{5}=\sin 75^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}, \\
d_{6}=1,
\end{array}
$$

Among them, $d_{1}, d_{2}, \cdots, d_{5}$ each have 12 segments, and $d_{6}$ has 6 segments.
Clearly, the combinations that cannot form a triangle are of the following four types:
$(2,2,6)$ type, with $\mathrm{C}_{12}^{2} \mathrm{C}_{6}^{1}$ ways;
$(1,1, k)(k \geqslant 3)$ type, with $\mathrm{C}_{12}^{2} \mathrm{C}_{42}^{1}$ ways;
$(1,2, k)(k \geqslant 4)$ type, with $\mathrm{C}_{12}^{1} \mathrm{C}_{12}^{1} \mathrm{C}_{30}^{2}$ ways;
$(1,3, k)(k \geqslant 5)$ type, with $\mathrm{C}_{12}^{1} \mathrm{C}_{12}^{1} \mathrm{C}_{30}^{2}$ ways.
Therefore, the required probability is $1-\frac{63}{286}=\frac{223}{286}$.