5. Intersecting Diagonals (from 7th grade. 2 points). In a convex hexagon, two random diagonals are chosen independently of each other. Find the probability that these diagonals intersect inside the hexagon (inside means not at a vertex).
Solution. A hexagon has a total of 9 diagonals, and thus $\frac{9 \cdot 8}{2}=36$ pairs of diagonals. It remains to determine how many pairs of diagonals intersect inside the hexagon. Diagonals in a hexagon can be of two types: main (connecting opposite vertices) and non-main. For example, diagonal $A C$ is non-main. There are a total of 6 non-main diagonals, and each intersects 3 other diagonals (see figure). There are 3 main diagonals (such as $A D$), and each intersects 4 other diagonals. In total, we get $\frac{6 \cdot 3 + 3 \cdot 4}{2}=15$ intersection points (the division by 2 is necessary because the expression in the numerator counts each intersection point twice).
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IX Olympiad in Probability Theory and Statistics for Schoolchildren. Correspondence Round.

From the Laboratory of Probability Theory and Statistics of the MCNMO, 2015-2016

Since the number of intersection points inside is the same as the number of pairs of diagonals intersecting inside, the probability that two randomly chosen diagonals intersect inside is

$$
\frac{15}{36}=\frac{5}{12}.
$$

Answer: $\frac{5}{12}$.