If three distinct vertices are chosen at rand | vedclass

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Question

(A) cube has $8$ vertices. The total number of ways to choose $3$ vertices out of $8$ is given by $^8C_3 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	ime

Vedclass · Accepted Answer

$1/7$

Vedclass · Answer

(A) cube has $8$ vertices. The total number of ways to choose $3$ vertices out of $8$ is given by $^8C_3 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56$.An equilateral triangle is formed by choosing $3$ vertices such that the distance between any two of them is equal to the length of the face diagonal of the cube.Each face of the cube has $2$ diagonals,and there are $6$ faces,but each diagonal is shared by $2$ faces. The total number of face diagonals is $12$.However,an equilateral triangle is formed by selecting $3$ vertices such that each pair is connected by a face diagonal. There are exactly $8$ such triangles in a cube (each corresponding to a set of $3$ vertices that are mutually equidistant).Therefore,the number of favorable outcomes is $8$.The probability is $\frac{8}{56} = \frac{1}{7}$.

If three distinct vertices are chosen at random from the vertices of a cube,then the probability that they form the vertices of an equilateral triangle is:

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