Question

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

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

Answer 1

The correct option is A $\frac{1}{10}$

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