7・7 Given six points $A, B, C, D, E, F$ chosen randomly on a given circumference, these points are chosen independently and are equally likely with respect to arc length. Find the probability that triangles $A B C$ and $D E F$ do not intersect (i.e., have no common points).
[Solution] Note that the number of cyclic permutations of six distinct points distributed on a circle is
$$
p_{5}^{5}=5!=120 \text {. }
$$

And due to the symmetry of the distribution, these permutations have the same probability. The number of different permutations where triangle $A B C$ and triangle $D E F$ do not intersect is
$$
p_{3}^{3} \cdot p_{3}^{3}=3!\times 3!=36 \text {. }
$$

This is determined by the internal order of $A, B, C$ and the internal order of $D, E, F$.
Therefore, the probability that triangle $A B C$ does not intersect with triangle $D E F$ is:
$$
p=\frac{36}{120}=\frac{3}{10} .
$$