Ten points are given in the plane and no three are collinear. Four distinct segments connecting pairs of these points are chosen at random, but all with the same probability. What is the probability that three of the chosen segments form a triangle?
The number of possible segments between the 10 points is $\binom{10}{2}=45$ and the number of ways to choose 4 of these segments is $\binom{45}{4}$.

The number of ways to choose 4 segments such that three of them form a triangle is equal to the number of ways to choose three vertices, which determine the three segments of the triangle, multiplied by the number of ways to choose the other segment, that is $\binom{10}{3} \times (45-3)$. Therefore, the probability that three of the four segments form a triangle is

$$
\frac{\binom{10}{3} \times 42}{\binom{45}{4}}=\frac{10 \times 9 \times 8 \times 42 \times 4!}{3! \times 45 \times 44 \times 43 \times 42}=\frac{16}{473}
$$