Six points are there on a circle from which two triangles drawn with no vertex common. Find the probability that none of the sides of the triangles intersect.

Explanation

Let the points on the circle be `A_(1), A_(2), A_(3), A_(4), A_(5)` and `A_(6)` in order.
Since two triangles are drawn with no vertex common and none of the sides of the triangles intersect, three vertices of each triangle are consecutive.
One such pair of triangles is `DeltaA_(1)A_(2)A_(3)` and `DeltaA_(4)A_(5)A_(6)`.
Similarly, we have pairs `(DeltaA_(2)A_(3)A_(4).DeltaA_(1)A_(5)A_(6)),...,(DeltaA_(6)A_(1)A_(2) and DeltaA_(3)A_(4)A_(5))`.
So, there are six favourable cases.
Also, total number of cases = `.^(6)C_(3)`
`therefore` Required probability = `(6)/(.^(6)C_(3)) = (3)/(10)`