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# Three distinct vertices are chosen at random form the vertices of a given regular polygon of (2n+1) sides. Let all such choices are equally likely and the probability that the centre of the given polygon lies in the interior of  the triangle determined by these three chosen random points is 5/14.Q. No. 1          The number of diagonals of the polygon is equal to                        (a) 14               (b) 18               (c) 20               (d) 27Q. No. 2          The number of points of intersection of the diagonals lying exactly inside the polygon is equal to                                    (a) 70               (b) 35               (c) 126             (d) 96Q. No. 3          There vertices of the polygon are chosen at random. The probability that these vertices from an isosceles triangle is                        (a) 1/3              (b) 3/7              (c) 3/28            (d) None of these

suyash sinha
13 Points
3 years ago
There are $\binom{2n+1}{3}$ ways how to pick the three vertices. We will now count the ways where the interior does NOT contain the center. These are obviously exactly the ways where all three picked vertices lie among some $n+1$ consecutive vertices of the polygon. We will count these as follows: We will go clockwise around the polygon. We can pick the first vertex arbitrarily ($2n+1$ possibilities). Once we pick it, we have to pick $2$ out of the next $n$ vertices ($\binom{n}{2}$ possibilities).Then the probability that our triangle does NOT contain the center is $p = \frac{ (2n+1){\binom{n}{2}} }{ {\binom{2n+1}{3} } } = \frac{ (1/2)(2n+1)(n)(n-1) }{ (1/6)(2n+1)(2n)(2n-1) } = \frac{ 3(n)(n-1) }{ (2n)(2n-1) }$And then the probability we seek is $1-p = \frac{ (2n)(2n-1) - 3(n)(n-1) }{ (2n)(2n-1) } = \frac{ n^2+n }{ 4n^2 - 2n } = \boxed{\frac{n+1}{4n-2}}$