Question

The sum of two positive quantities is equal to $2n$. The probabilities that their product is not less than $\frac{3}{4}$ times their product is

• A. $\frac{3}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. none of these

Answer 1

The correct option is B $\frac{1}{2}$

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 $=\frac{3n}{2}-\frac{n}{2}=n.$

Hence, required probability $=\frac{n}{2n}=\frac{1}{2}.$