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 :
A.
$$\frac{1}{{20}}$$
B.
$$\frac{{11}}{{20}}$$
C.
$$\frac{1}{3}$$
D.
$$\frac{3}{{20}}$$
Answer :
$$\frac{{11}}{{20}}$$
Solution :
Given equation
$$\eqalign{
& x + \frac{{100}}{x} > 50 \cr
& \Rightarrow {x^2} - 50x + 100 > 0\,\, \Rightarrow {\left( {x - 25} \right)^2} > 525 \cr
& \Rightarrow x - 25 < - \sqrt {\left( {525} \right)} {\text{ or }}x - 25 > \sqrt {\left( {525} \right)} \cr
& \Rightarrow x < 25 - \sqrt {\left( {525} \right)} {\text{ or }}x > 25 + \sqrt {\left( {525} \right)} \cr} $$
As $$x$$ is positive integer and $$\sqrt {\left( {525} \right)} = 22.91,$$ we must have
$$x \leqslant 2{\text{ or }}x \geqslant 48$$
Let $$E$$ be the event for favourable cases and $$S$$ be the sample space.
$$\eqalign{
& \therefore \,E = \left\{ {1,\,2,\,48,\,49,\,......,\,100} \right\} \cr
& \therefore \,n\left( E \right) = 55{\text{ and }}n\left( S \right) = 100 \cr} $$
Hence the required probability
$$P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{55}}{{100}} = \frac{{11}}{{20}}$$