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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
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$A=\left\{n \in S: n+\frac{50}{n}>27\right\}, B=\{n \in S: n$ is a prime) and $C=\{n \in S: n$ is a square). Then, correct order of their probabilities is
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Verified Answer
The correct answer is:
$P(A)>P(B)>P(C)$
Given that $S=\{1,2,3 \ldots, 50\}$
$\begin{aligned}
& A=\left\{n \in S: n+\frac{50}{n}>27\right\} \\
& =\{n \in S: n < 2 \text { or } n>25\} \\
& =\{1,26,27, \ldots, 50\} \\
& \Rightarrow \quad n(A)=26 \\
& B=\{n \in S: n \text { is a prime }\} \\
& =\{2,3,5,7,11,13,17,19,23,29 \text {, } \\
& 31,37,41,43,47\} \\
& \Rightarrow \quad n(B)=15 \\
& C=\{n \in S: n \text { is a square }\} \\
& =\{1,4,9,16,25,36,49\} \\
& \Rightarrow \quad n(C)=7 \\
& \therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{26}{50} \\
& P(B)=\frac{n(B)}{n(S)}=\frac{15}{50} \\
& P(C)=\frac{n(C)}{n(S)}=\frac{7}{50} \\
& \Rightarrow \quad P(A)>P(B)>P(C) \\
&
\end{aligned}$
$\begin{aligned}
& A=\left\{n \in S: n+\frac{50}{n}>27\right\} \\
& =\{n \in S: n < 2 \text { or } n>25\} \\
& =\{1,26,27, \ldots, 50\} \\
& \Rightarrow \quad n(A)=26 \\
& B=\{n \in S: n \text { is a prime }\} \\
& =\{2,3,5,7,11,13,17,19,23,29 \text {, } \\
& 31,37,41,43,47\} \\
& \Rightarrow \quad n(B)=15 \\
& C=\{n \in S: n \text { is a square }\} \\
& =\{1,4,9,16,25,36,49\} \\
& \Rightarrow \quad n(C)=7 \\
& \therefore \quad P(A)=\frac{n(A)}{n(S)}=\frac{26}{50} \\
& P(B)=\frac{n(B)}{n(S)}=\frac{15}{50} \\
& P(C)=\frac{n(C)}{n(S)}=\frac{7}{50} \\
& \Rightarrow \quad P(A)>P(B)>P(C) \\
&
\end{aligned}$
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