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Q.
A natural number x is chosen at random from the first 100 natural numbers. Then the probability, for the equation $x+\frac{100}{x} > 50$ to be true is
Probability
Solution:
Given equation
$x+\frac{100}{x} > 50 \Rightarrow x^{2}-50x+100 > 0 \Rightarrow \left(x-25\right)^{2} > 525$
$\Rightarrow x-25 <-\sqrt{\left(525\right)}$ or $x-25 > \sqrt{\left(525\right)}$
$\Rightarrow x < 25-\sqrt{\left(525\right)}$ or $x > 25+\sqrt{\left(525\right)}$
As x is positive integer and $\sqrt{\left(525\right)}=22.91$, we must have $x\le2$ or $x \ge48$
Let E be the event for favourable cases and S be the sample space.
$\therefore E=\left\{1, 2, 48, 49, ......100\right\}$
$\therefore n\left(E\right)= 55$ and $n\left(S\right) =100$
Hence the required probability $P\left(E\right) =\frac{n\left(E\right)}{n\left(S\right)}=\frac{55}{100}=\frac{11}{20}$