Question

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

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{2}{3}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Let one of the number (quantity)be x.So the other is 2n-x.
Now product of x and 2n-x will be largest if these two numbers are equal
$\therefore 2n-x=x$
$\therefore x=n$
$\therefore$ both are equal to n so product $=n^2$
Now by problem $x(2n-x)\ge \frac{3}{4}\times$ their product
Note n(S)=2n $\Rightarrow x(2n-x)\ge \frac{3}{4}{n}^2$
$\Rightarrow 4x^2-8nx+3n^2\le 0$ $\Rightarrow \frac{n}{2}\le x\le \frac{3}{2}n$
Now $n\left(A\right)=\frac{3n}{2}-\frac{n}{2}=n$
$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{n}{2n}=\frac{1}{2}$