in n sided regular hexagon find the probability that the two diagonals chosen at random will intersect inside the polygon.?

the number of diagonals = ${}^{6}C_{2}-6=\frac{6*5}{2}-6=9$

now the number of ways of choosing any 2 diagonals = ${}^{9}C_{2}=\frac{9*8}{2}=36$

now since the diagonals intersect inside the polygon, let us choose any 4 points out of 6 vertices,

so that they will form a quadrilateral and the diagonals of a quadrilateral must intersect inside it.

thus the total number of ways such that diagonals will intersect inside the hexagon = ${}^{6}C_{4}=\frac{6*5}{2}=15$

thus the required probability = 15/36 = 5/12

hope this helps you

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