If three of the six vertices of a regular hex | vedclass

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

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$\frac{1}{10}$

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(C) The total number of ways to choose $3$ vertices out of $6$ is given by $^{6}C_{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.In a regular hexagon,an equilateral triangle is formed by choosing vertices that are separated by one vertex each. The possible equilateral triangles are $	riangle A_{1}A_{3}A_{5}$ and $	riangle A_{2}A_{4}A_{6}$.Thus,there are $2$ such equilateral triangles.The probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{2}{20} = \frac{1}{10}$.

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

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