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

in a hexagon , number of sides = 6
the number of diagonals = C26-6=6*52-6=9
now the number of ways of choosing any 2 diagonals = C29=9*82=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 = C46=6*52=15
thus the required probability = 15/36 = 5/12

hope this helps you

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