Question

A natural number $x$ is chosen at random from the first one hundred natural numbers. The probability that $\frac{(x-20)(x-40)}{(x-30)}<0$ is

• A. $\frac{1}{50}$
• B. $\frac{3}{50}$
• C. $\frac{7}{25}$
• D. $\frac{9}{50}$

Answer 1

The correct option is C $\frac{7}{25}$

$x\in N,$ Total ways ${}^{100}{C}_1=100$
$\Rightarrow n\left(S\right)=100$
$\frac{(x-20)(x-40)}{(x-30)}<0\therefore x\in (-\infty ,20)\cup (30,40)$
$\therefore E=\{1,2,3,...19,31,32,33,...,39\}$
$\Rightarrow n\left(E\right)=28$
$\therefore$ Required probability $=\frac{28}{100}=\frac{7}{25}$