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Three vertices are chosen randomly from the nine vertices of a regular 9 -sided polygon. The probability that they form the vertices of an isosceles triangle, is
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Verified Answer
The correct answer is:
$\frac{3}{7}$
Number of triangles formed $={ }^9 C_3$
Number of isosceles triangles $=9 \times\left(\frac{9-1}{2}\right)$
$=9 \times 4=36$
So, required probability
$=\frac{36}{{ }^9 C_3}=\frac{36}{\frac{9 !}{3 !(9-3) !}}=\frac{36 \times 3 \times 2 \times 6 !}{9 \times 8 \times 7 \times 6 !}=\frac{3}{7}$
Number of isosceles triangles $=9 \times\left(\frac{9-1}{2}\right)$
$=9 \times 4=36$
So, required probability
$=\frac{36}{{ }^9 C_3}=\frac{36}{\frac{9 !}{3 !(9-3) !}}=\frac{36 \times 3 \times 2 \times 6 !}{9 \times 8 \times 7 \times 6 !}=\frac{3}{7}$
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