Question

In a convex polygon of $6$ sides two diagonals are selected at random. The probability that they intersect at an interior point of the polygon is:

• A. $\frac{2}{5}$
• B. $\frac{5}{12}$
• C. $\frac{7}{12}$
• D. $\frac{3}{5}$

Answer 1

The correct option is B $\frac{5}{12}$

Total number of diagonals in the polygon = ${}^6{C}_2-6=9$

Number of ways of selecting 2 out of 9 diagonals (Total ways) = ${}^9{C}_2=36$

Now note that you can form a quadrilateral by selecting any 4 vertices out of 6. And each quadrilateral will have a pair of intersecting diagonals. So favorable ways = ${}^6{C}_4=15$

Hence the required probability = $\frac{\text{number of favorable ways}}{\text{Total number of ways}}=\frac{15}{36}=\frac{5}{12}$