Question

From the set of natural numbers first thirty are chosen randomly, then the probability that $x+\frac{300}{x}>40$ is

• A. $\frac{3}{10}$
• B. $\frac{2}{3}$
• C. $\frac{1}{30}$
• D. $\frac{5}{6}$

Answer 1

The correct option is A $\frac{3}{10}$

Here, $n\left(S\right)=30$
Now $x+\frac{300}{x}>40\Rightarrow x^2-40x+300>0\Rightarrow (x-10)(x-30)>0$
$\Rightarrow x<10\text{or}x>30$
Let $E$ is the favorable case $\therefore n\left(E\right)=n\{1,2,3,4,5,6,7,8,9\}=9$
Hence required probability is $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{10}$