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).

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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 a total of three main diagonals (such as $A D$), and each intersects four other diagonals. In total, (6$\cdot$3 + 3$\cdot$4) : 2 = 15 pairs of intersecting diagonals.

![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-43.jpg?height=386&width=1092&top_left_y=1&top_left_x=486)

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

## Answer