Question

Given that the sum of two non-negative integers is $200$, the probability that their product is not less that $\frac{3}{4}$ times their greatest product value is

• A. $\frac{7}{16}$
• B. $\frac{101}{200}$
• C. $\frac{9}{16}$
• D. None of these

Answer 1

The correct option is B $\frac{101}{200}$

Let two quantities be $x$ and $y$.

Since the sum of two non-negative quantities is fixed, the product will be maximum when they are equal.

So, the greatest product $=xy=10000,$ where $x=y=100.$

Now,$xy\nless \left(\frac{3}{4}\right)\times 10000\Rightarrow xy\ge \left(\frac{3}{4}\right)\times 10000$

$\Rightarrow xy\ge 7500\Rightarrow x(200-x)\ge 7500$

$\Rightarrow x^2-200x+7500\le 0$

$\Rightarrow (x-50)(x-150)\le 0\Rightarrow 50\le x\le 150$

Thus, the favourable number of cases $=150-50+1=101$

Total number of ways $=200$

Hence, required probability $=\frac{101}{200}$