Three out of 6 vertices of a regular hexagon | vedclass

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(C) Given that $3$ out of $6$ vertices of a regular hexagon are chosen.Total number of ways to choose $3$ vertices out of $6$ is given by $^6C_3 = \fr

Vedclass · Accepted Answer

$\frac{1}{10}$

Vedclass · Answer

(C) Given that $3$ out of $6$ vertices of a regular hexagon are chosen.Total number of ways to choose $3$ vertices out of $6$ is given by $^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.In a regular hexagon,only two equilateral triangles can be formed by joining its vertices,which are $	riangle ACE$ and $	riangle BDF$.Thus,the number of favorable outcomes is $2$.Therefore,the probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{2}{20} = \frac{1}{10}$.

Three out of $6$ vertices of a regular hexagon are chosen at random. The probability that the triangle formed with these three vertices is an equilateral triangle is

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