$\left.\begin{array}{l}\text { [Discrete distribution] } \\ \text { [combinations and permutations] }\end{array}\right]$

In a convex hexagon, two random diagonals are independently chosen.

Find the probability that these diagonals intersect inside the hexagon (inside means not at a vertex).
A hexagon has a total of nine diagonals, and thus $9 \cdot 8: 2=36$ 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 a non-main diagonal (see figure). There are a total of 6 non-main diagonals, and each intersects 3 other diagonals. There are three main diagonals (such as $A D$), and each intersects four other diagonals. In total, we get (see figure) (6$\cdot$3 + 3$\cdot$4) : 2 = 15 pairs of intersecting diagonals.
![](https://cdn.mathpix.com/cropped/2024_05_06_e8dfada95fe97b3690e8g-04.jpg?height=382&width=1082&top_left_y=0&top_left_x=487)

Therefore, the probability that two randomly chosen diagonals intersect inside is $15 / 36=$ $5 / 12$.

## Answer

$5 / 12$.