In a convex hexagon two diagonals are drawn at random. The probability that the diagonals intersect at an interior point of the hexagon is
Answer & explanation
Correct answer: option 1
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 = $\frac{15}{36}=\frac{5}{12}$