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(D) The total number of ways to choose $3$ vertices out of $7$ is given by $^7C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.For a regula

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$\frac{3}{5}$

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(D) The total number of ways to choose $3$ vertices out of $7$ is given by $^7C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.For a regular $7$-sided polygon,consider a vertex $A$. We can form isosceles triangles with $A$ as the apex by choosing pairs of vertices that are equidistant from $A$ along the perimeter. For each vertex,there are $3$ such pairs of vertices,forming $3$ isosceles triangles.Since there are $7$ vertices in total,the total number of isosceles triangles is $7 	imes 3 = 21$.Therefore,the required probability is $\frac{21}{35} = \frac{3}{5}$.

Three vertices are chosen randomly from the seven vertices of a regular $7$-sided polygon. The probability that they form the vertices of an isosceles triangle is

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