A number n is chosen at random from S={1, 2, | vedclass

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(B) Given $S=\{1, 2, 3, \ldots, 50\}$,so $n(S) = 50$. For set $A$,we solve $n + \frac{50}{n} > 27$. Multiplying by $n$ (since $n > 0$),we get $n^2 - 2

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$P(A) > P(B) > P(C)$

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(B) Given $S=\{1, 2, 3, \ldots, 50\}$,so $n(S) = 50$. For set $A$,we solve $n + \frac{50}{n} > 27$. Multiplying by $n$ (since $n > 0$),we get $n^2 - 27n + 50 > 0$. Factoring the quadratic,$(n-25)(n-2) > 0$. This holds for $n  25$. Since $n \in S$,$n=1$ or $n \in \{26, 27, \ldots, 50\}$. Thus,$A = \{1, 26, 27, \ldots, 50\}$,so $n(A) = 1 + 25 = 26$. For set $B$,the primes in $S$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$,so $n(B) = 15$. For set $C$,the squares in $S$ are $\{1, 4, 9, 16, 25, 36, 49\}$,so $n(C) = 7$. Calculating probabilities: $P(A) = \frac{26}{50}$,$P(B) = \frac{15}{50}$,$P(C) = \frac{7}{50}$. Therefore,$P(A) > P(B) > P(C)$.

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

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