Three of the six vertices of a regular hexago | vedclass

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(C) The total number of ways to choose $3$ vertices out of $6$ is given by the combination formula ${}^6C_3 = \frac{6 	imes 5 	imes 4}{1 	imes 2

Vedclass · Accepted Answer

$1/10$

Vedclass · Answer

(C) The total number of ways to choose $3$ vertices out of $6$ is given by the combination formula ${}^6C_3 = \frac{6 	imes 5 	imes 4}{1 	imes 2 	imes 3} = 20$. In a regular hexagon,there are exactly $2$ equilateral triangles that can be formed by connecting the vertices. These are formed by taking alternating vertices (e.g.,vertices $1, 3, 5$ and $2, 4, 6$). Therefore,the required probability is $\frac{	ext{Number of equilateral triangles}}{	ext{Total number of triangles}} = \frac{2}{20} = \frac{1}{10}$.

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

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