A natural number x is chosen at random from the first $100$ natural numbers. The probability for the inequality $x+\frac{100}{x}>50$ to hold good is:

The correct option is D $\frac{11}{20}$
Given: $x+\frac{100}{x}>50$
$\Rightarrow x^2-50x+100>0$
$\Rightarrow (x-25{)}^2>525\Rightarrow x-25<-\sqrt{525}$ or $x-25>\sqrt{525}$
As $\sqrt{525}=22.91$ we must have $x\le 2$or$x\ge 48$
Let $E$ be the event for favorable cases and $S$ be the sample space.
$\therefore$ $E=1,2,48,49,\cdots 100$
$n\left(E\right)=55,n\left(S\right)=100$
$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{55}{100}=\frac{11}{20}$