Question

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

• A. $1/2$
• B. $1/5$
• C. $1/10$
• D. $1/20$

Answer 1

Answer: (C)

$1/10$

$n\left(S\right)=$ 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 $=\frac{2}{20}=\frac{1}{10}$