Question

The sum of two postive quantities is equal to $2n.$ The probability that their product is not less than $\frac{3}{4}$ times their greatest product is $\frac{k}{8}$ then find $k$

Answer 1

Let one of the quantities be $x.$ Thus the other is $2n-x.$
Then product will be greatest when they are equal $i.e.,$ each is $n$ in which case the product is $n^2.$
By the given condition
$x(2n-x)\ge \frac{3}{4}{n}^2$
$\Rightarrow 8nx-{4x}^2\ge {3n}^2\Rightarrow {4x}^2-8nx+{3n}^2\le 0$
$\Rightarrow (2x-3n)(2x-n)\le 0\Rightarrow \frac{n}{2}\le x\le \frac{3}{2}n.$
$\therefore$ favourable number of cases $=\frac{3n}{2}-\frac{n}{2}$
Hence required probability $=\frac{n}{2n}=\frac{1}{2}$