Inequality in Probability

Note that $x > 0$ since $x \in \{1, 2, 3 \ldots 100\}$ . Thus, we can multiply $x$ on both sides without changing the sign of the inequality. We have

$\displaystyle \begin{aligned} x + \frac{100}{x} & > 50 \\ x^2 + 100 & > 50x \\ x^2 - 50x + 100 & > 0 \end{aligned}$

Using the quadratic formula, we find that the roots of $f(x) = x^2 - 50x + 100$ are $x = 25 \pm 5\sqrt{21}$ , which are approximately $2.09$ and $47.91$ , respectively. We test the intervals $0 < x < 2.09$ , $2.09 < x < 47.91$ , and $47.91 < x < 100$ , using $x = 2 < 2.09$ , $x = 3 > 2.09$ , and $x = 48 > 47.91$

$f(2) = 2^2 - 50(2) + 100 = 4 \ {\color{#3D99F6}{> 0}} \\ \implies 0 < x \leq \ 2$

$f(3) = 3^2 - 50(3) + 100 = -41 \ {\color{#D61F06}{< 0}} \\ \implies 3 \cancel{\leq} x \cancel{<} 48$

$f(48) = 48^2 - 50(48) + 100 = 4 \ {\color{#3D99F6}{> 0}} \\ \implies 48 \leq x \leq \ 100$

Our solution set is $\{1, 2, 48, 49, 50 \ldots 100\}$ , which consists of $55$ integers. Thus, the probability that $x + \dfrac{100}{x} > 50$ for the first $100$ natural numbers is $\dfrac{55}{100} = \boxed{\dfrac{11}{20}}$ .