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(C) The total number of ways to choose $3$ vertices out of $6$ to form a triangle is given by the combination formula $n = \binom{6}{3} = \frac{6 	im

Vedclass · Accepted Answer

$1/10$

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(C) The total number of ways to choose $3$ vertices out of $6$ to form a triangle is given by the combination formula $n = \binom{6}{3} = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$. In a regular hexagon,there are exactly $2$ equilateral triangles that can be formed by joining its vertices (these are formed by taking alternate vertices). Therefore,the number of favorable outcomes is $r = 2$. The probability of the event is $P = \frac{r}{n} = \frac{2}{20} = \frac{1}{10}$.

If three vertices are chosen from the six vertices of a regular hexagon and joined to form a triangle,what is the probability that the triangle formed is an equilateral triangle?

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