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Question

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
$\frac{1}{7}$
$\frac{1}{3}$
$\frac{3}{7}$
$\frac{3}{5}$

A

\frac{1}{7}

B

\frac{1}{3}

C

\frac{3}{7}

D

\frac{3}{5}

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Solution

Verified by Toppr

A regular 3 sided polygon is nothing else but a heptagon for creating isosceles triangle we need to choose adjacent sides only.
No. of $△^{'} s$ formed $= 7 C_{3}$
While number of isosceles triangle formed $=$ No. of points $\times$ points available $= 7 \times 3.$
$\Rightarrow$ So, probability $= \frac{7 \times 3}{7 C_{3}} = \frac{21}{35} = \frac{3}{5} .$
Hence, the answer is $\frac{3}{5} .$

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