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(C) The total number of ways to choose $3$ vertices out of $6$ is given by $n(S) = {}^{6}C_{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$

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$1/10$

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(C) The total number of ways to choose $3$ vertices out of $6$ is given by $n(S) = {}^{6}C_{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$. An equilateral triangle is formed if we select alternate vertices of the regular hexagon. In a regular hexagon with vertices labeled $1, 2, 3, 4, 5, 6$,the possible sets of vertices forming an equilateral triangle are $\{1, 3, 5\}$ and $\{2, 4, 6\}$. Thus,the number of favorable outcomes is $n(A) = 2$. The probability $P(A)$ is given by $P(A) = \frac{n(A)}{n(S)} = \frac{2}{20} = \frac{1}{10}$.

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

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