Question

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

• A. $\frac{5}{12}$
• B. $\frac{7}{12}$
• C. $\frac{2}{5}$
• D. None of these

Answer 1

Answer: (A)

$n(S)=$ total number of selections of two diagonals
$={}^9{C}_2=\frac{9\times 8}{2}=36$
$n(E)=$ the number of selections of two diagonals which intersect at an interior point $=$ the total number of selections of four vertices $={}^6{C}_4=15$
$\therefore$ The required probability $=\frac{15}{36}=\frac{5}{12}$.