Three of the six vertices of a regular hexago | vedclass

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Question

(C) regular hexagon has $6$ vertices. The total number of ways to choose $3$ vertices out of $6$ is given by the combination formula ${}^nC_r = \frac{

Vedclass · Accepted Answer

$\frac{1}{10}$

Vedclass · Answer

(C) regular hexagon has $6$ vertices. The total number of ways to choose $3$ vertices out of $6$ is given by the combination formula ${}^nC_r = \frac{n!}{r!(n-r)!}$.Total triangles $= {}^6C_3 = \frac{6 	imes 5 	imes 4}{3 	imes 2 	imes 1} = 20$.In a regular hexagon,an equilateral triangle is formed by connecting every second vertex. There are only $2$ such triangles possible (by taking vertices ${1, 3, 5}$ and ${2, 4, 6}$).Therefore,the required probability $= \frac{	ext{Number of equilateral triangles}}{	ext{Total number of triangles}} = \frac{2}{20} = \frac{1}{10}$.

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:

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