Probability Question 350
Question: 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 $ is
Options:
A) $ \frac{1}{20} $
B) $ \frac{11}{20} $
C) $ \frac{1}{3} $
D) $ \frac{3}{20} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given equation $ 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)} $
$ \Rightarrow x<25-\sqrt{(525)} $ or $ x>25+\sqrt{(525)} $
As x is positive integer and $ \sqrt{(525)}=22.91, $ we must have
$ x\le 2 $ or $ x\ge 48 $
Let E be the event for favourable cases and S be the sample space.
$ \therefore E={1,2,48,49,…100} $
$ \therefore n(E)=55 $ and $ n(S)=100 $
Hence the required probability
$ P(E)=\frac{n(E)}{n(S)}=\frac{55}{100}=\frac{11}{20} $
BETA
