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Using the vertices of a polygon having 12 sides a triangle is constructed at random. The probability that the triangle so formed is such that no side of the polygon is side of the triangle is

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$\frac{28}{55}$

$\frac{7}{55}$

$\frac{17}{55}$

$\frac{18}{55}$

answer is B.

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Detailed Solution

$n (S) =^{12} C_{3}$

E= triangle so formed using verties no side of the polygon in side of the triangle.

$n (E) = {}^{n}C_{3} - n - n (n - 4)$ where $n = 12$

$P (E) = \frac{^{12} C_{3} - 12 - 12 \times 8}{^{12} C_{3}}$

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