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

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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

Aicharya · Accepted Answer

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

A number $n$ is chosen at random from $S=\{1,2,3, \ldots, 50\}$. Let $A=\left\{n \in S:…

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