Question

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

Answer 1

We choose three vertices out of 6 in ⁶C₃ = 20 ways.
∴ Total number of elementary events = n(S) = 20
Number of ways of choosing an equilateral triangle = 2

i.e. A₁A₃A₅ or A₂A₄A₆
∴ Total number of favourable events = n(E) = 2
Hence, required probability = \[\frac{2}{20} = \frac{1}{10}\]