Question

A number $n$ is chosen at random from $S=\{1,2,3,\dots ,50\}.$ 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 $S=\{1,2,3\dots ,50\}$
$A=\left\{n\in S:n+\frac{50}{n}>27\right\}$
$=\{n\in S:n<2$ or $n>25\}$
$=\{1,26,27,\dots ,50\}$
$\Rightarrow n(A)=26$
$B=\{n\in S:n$ 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 $\}$
$=\{1,4,9,16,25,36,49\}$
$\Rightarrow n(C)=7$
$\therefore P(A)=\frac{n(A)}{n(S)}=\frac{26}{50}$
$P(B)=\frac{n(B)}{n(S)}=\frac{15}{50}$
$P(C)=\frac{n(C)}{n(S)}=\frac{7}{50}$
$\Rightarrow P(A)>P(B)>P(C)$