27.3. Let's divide all possible pairs of triples of vertices into $C_{n}^{\natural}$ groups, collecting in one group those and only those pairs of triples that form the same set of six vertices. On the one hand, each such group contains as many elements as there are ways to split a fixed set of six vertices into two triples, i.e., $C_{b}^{s}=20$ elements. On the other hand, there are exactly 6 ways to split a set of six vertices into two triples that satisfy the required condition in the problem. Therefore, the desired probability is $6 / 20=0.3$.
27.3. (СІІІА, 83). From the vertices of a regular $n$-gon $(n \geqslant 6)$, two different triples of points are chosen at random. What is the probability that the two triangles, whose vertices are the chosen triples, do not intersect?