Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{5}$
• C. $\frac{1}{10}$
• D. $\frac{1}{20}$

Answer 1

Answer: (C)

$\frac{1}{10}$

$n\left(S\right)=$ Number of ways of selection of 3 vertices out of 6 ${=}^6{C}_3=20$
Out of which there will be only two cases where triangle will be equilateral (By selecting ACE or BDF, where A,B,C,D,E,F are vertices of hexagon).
Hence required probability $=\frac{2}{20}=\frac{1}{10}$