in n sided regular hexagon find the probability that the two diagonals chosen at random will intersect inside the polygon.?

in a hexagon , number of sides = 6
the number of diagonals = ${}^{6}C_2-6=\frac{6*5}{2}-6=9$
now the number of ways of choosing any 2 diagonals = ${}^{9}C_2=\frac{9*8}{2}=36$
now since the diagonals intersect inside the polygon, let us choose any 4 points out of 6 vertices,
so that they will form a quadrilateral and the diagonals of a quadrilateral must intersect inside it.
thus the total number of ways such that diagonals will intersect inside the hexagon = ${}^{6}C_4=\frac{6*5}{2}=15$
thus the required probability = 15/36 = 5/12

hope this helps you