In a convex hexagon two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the hexagon is

$\frac{5}{12}$

Number of diagonals of a hexagon = ${^6C}_2 −6=9$

Number of ways of selecting two diagonals = ${^6C}_2 = 36$

Number of ways of selecting two intersecting diagonals = Number of ways of selecting four vertices of the hexagon = ${^6C}_4 =15$

Hence, required probability = $\frac{15}{36}=\frac{5}{12}$