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

We have, Number of diagonals of a hexagon $={}^{6}C_2-6=9$

$\therefore n\left(s\right)=$ Total number of selections of two diagonals $={}^{9}C_2=36$

and $n\left(E\right)=$ The number of selections of two diagonals which intersect at an interior point

$=$ The number of selections of four vertices $={}^{6}C_4=15$

Hence, required probability $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{15}{36}=\frac{5}{12}$