In a n-sided regular polygon,the probability that the two diagonal chosen at random will intersect inside the polygon is

\frac{2^{2} C_{2}}{^{(^{n} C_{2} - n)} C_{2}}

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\frac{^{n (n - 1)} C_{2}}{^{(^{n} C_{2} - n)} C_{2}}

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\frac{^{n} C_{4}}{^{(^{n} C_{2} - n)} C_{2}}

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Solution

The correct option is C $\frac{^{n} C_{4}}{^{(^{n} C_{2} - n)} C_{2}}$
In $n$ sided polygon numbers of diagonals $=^{n} C_{2} - n$
Number of ways of choosing any 2 diagonal $= {^{(^{n} C_{2} - n)} C}_{2}$
Now out of these $n$ sides we shall choose $4$ points so that they will form a quadrilateral and diagonal must intersect inside it.
No. of ways $=^{n} C_{4}$

$∴$ probability $= \frac{^{n} C_{4}}{{^{(^{n} C_{2} - n)} C}_{2}}$
Hence, the answer is $\frac{^{n} C_{4}}{{^{(^{n} C_{2} - n)} C}_{2}} .$

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