Question

A number $n$ is chosen at random from $S=\{1,2,3, \ldots, 50\} .$ Let
$A=\left\{n \in S: n+\frac{50}{n}>27\right\}, B=\{n \in S: n$
is a prime ) and $C=\{n \in S: n$ is a square). Then, correct order of their probabilities is

• A. $P(A) < P(B) < P(C)$
• B. $P(A) > P(B) > P(C)$
• C. $P(B) < P(A) < P(C)$
• D. $P(A) > P(C) > P(B)$

Answer 1

Answer: (B)

Given that $S=\{1,2,3 \ldots, 50\}$
$A =\left\{n \in S: n+\frac{50}{n}>27\right\} $
$=\{n \in S: n < 2$ or $ n >25\} $
$=\{1,26,27, \ldots, 50\}$
$\Rightarrow n(A)=26$
$B=\{n \in S: n$ is a prime $\}$
$=\{2,3,5,7,11,13,17,19,23,29,31, 37, 41, 43, 47\}$
$\Rightarrow n(B)=15 $
$C =\{n \in S: n$ is a square $\}$
$=\{1,4,9,16,25,36,49\} $
$ \Rightarrow n(C)=7 $
$\therefore P(A) =\frac{n(A)}{n(S)}=\frac{26}{50} $
$P(B)=\frac{n(B)}{n(S)}=\frac{15}{50}$
$P(C)=\frac{n(C)}{n(S)}=\frac{7}{50}$
$ \Rightarrow P(A) > P(B) > P(C) $