Question

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

• A. $\frac{1}{6}$
• B. $\frac{1}{3}$
• C. $\frac{1}{10}$
• D. $noneofthese$

Answer 1

Answer: (C)

$\frac{1}{10}$

There are 6 vertices in a hexagon. Using 3 vertices out of 6 vertices we can form ${}^6{C}_3$ triangles. But there can be only two triangles out of ${}^6{C}_3$ triangles which are equilateral (see the figure, (i) $\Delta$ACE, (ii) $\Delta$ BDF)

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