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

We can choose three vertices out of 6 in ${ }^6{\mathrm{C}}_3=20$  ways.
Chosen vertices can form an equilateral triangle in just two ways ${\mathrm{A}}_1{\mathrm{A}}_3{\mathrm{A}}_5\text{ and }{\mathrm{A}}_2{\mathrm{A}}_4{\mathrm{A}}_6$
.'. Required probability $=\frac{2}{20}=\frac{1}{10}$