Example 7 (1983 USA Mathematical Olympiad) On a given circle, six points $A, B, C, D, E, F$ are chosen randomly. The choices are independent and equally likely with respect to arc length. Find the probability that $\triangle A B C$ and $\triangle D E F$ do not intersect (i.e., have no common points).

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Note that the number of cyclic permutations of six distinct points distributed on a circle is
$$
A_{n}^{5}=5!=120 .
$$

And due to the symmetry of the distribution, these permutations have the same probability, where the number of different arrangements in which $\triangle A B C$ and $\triangle D E F$ do not intersect is $A_{3}^{3} \cdot A_{3}^{3}=3!\times 3!=36$.
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:
$$
P=\frac{36}{120}=\frac{3}{10} .
$$