If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is

Choosing three vertices of a regular hexagon alternately, only two equilateral triangles are possible,

i.e., $A_1,A_3,A_5$ and $A_2,A_4,A_6$ Hence,

required probability= $=\frac{2}{{ }^6{C}_3}=\frac{2}{20}=\frac{1}{10}$