A number $\mathrm{n}$ is chosen at random from $s=\{1,2,3, \ldots, 50\}$. Let $\mathrm{A}=\{n \in s: n$ is a square $\}$, $\mathrm{B}=\{n \in s: n$ is a prime $\}$ and $\mathrm{C}=\{n \in s: n$ is a square $\}$. Then, correct order of their probabilities is

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Aicharya · Accepted Answer

Given, $S=\{1,2,3 \ldots, 50\}$ $\begin{aligned} A & +\left\{n \in S: n+\frac{50}{n}>27ight\} \ & =\left\{n \in S: n^2-27 n+50>0ight\} \ & =\{n \in S:(n-25)(n-2)>0\} \ & =\{n \in S: n 25\} \ & =\{1,26,27,28, \ldots, 50\} \end{aligned}$ $\begin{aligned} & \Rightarrow n(A)=26 \ & B=\{n \in S: n 	ext { is prime }\} \ & =\{2,3,5,7,11,13,17,19,23 	ext {, } \ & 29,31,37,41,43,47\} \ & \Rightarrow n(B)=15 \ & C=\{n \in S: n 	ext { is a square }\}=\{1,4,9,16,25,36,49\} \ & \Rightarrow n(C)=7 \ & 	herefore \quad p(A)=\frac{n(A)}{n(S)}=\frac{26}{50} 	ext {, } \ & \Rightarrow \quad p(B)=\frac{n(B)}{n(S)}=\frac{15}{50} 	ext {, } \ & p(C)=\frac{n(C)}{n(S)}=\frac{7}{50} \ & 	herefore \quad p(A)>p(B)>p(C) \ & \end{aligned}$

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

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