Question

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals :

(a) \(\frac{1}{6}\)

(b) \(\frac{1}{3}\)

(c) \(\frac{1}{10}\)

(d) None of these

Answer 1

(c) \(\frac{1}{10}\)

Let S be the sample space.

Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon

= ⁶C₃ = \(\frac{6\times5\times4}{3\times2}\) = 20.

Let A : Event that the selected three vertices form an equilateral triangle. 

Then n(A) = 2 

(As only two equilateral triangles are formed from the vertices of a regular hexagon)

∴ Required probability = \(\frac{n(A)}{n(S)}\) = \(\frac{2}{20}\) = \(\frac{1}{10}\).