Question

Let $x+y=20,x,y\in W$, the set of non-negative integers. If $S$ is the maximum value of $xy$ and the probability of $xy$ not less than $\frac{3S}{4}$ is $\frac{n}{m}$, where $m$ and $n$ are co-prime, then the value of $n+m$ is

Answer 1

Given $x+y=20$
The maximum value of $xy$ is when $x=y=10$,
$S=10\times 10=100\Rightarrow \frac{3S}{4}=75$

Now, $xy\ge 75$
$\Rightarrow x(20-x)\ge 75\Rightarrow 20x-x^2\ge 75\Rightarrow x^2-20x+75\le 0\Rightarrow (x-5)(x-15)\le 0\Rightarrow x\in [5,15]$
Required number of values of $x$ is $11$
Total number of possible values of $x$ is $20$
Therefore, the required probability $=\frac{11}{20}$
$\Rightarrow \frac{n}{m}=\frac{11}{20}\Rightarrow n+m=31$