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18 Apr 2026, 00:41
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14% (03:56) correct
86%
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\(S\) is the set of the \(n\) smallest positive integers, where \(n > 5\). If two distinct integers are chosen at random from \(S\), the probability that the larger of the two integers is equal to \(n - 2\) is \(\frac{1}{6}\). What is the probability that the larger of the two integers chosen is greater than \(n - 2\)?
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{5}{12}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{5}{6}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{5}{12}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{5}{6}\)
18 Apr 2026, 12:50
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\(S\) is the set of the \(n\) smallest positive integers, where \(n > 5\). If two distinct integers are chosen at random from \(S\), the probability that the larger of the two integers is equal to \(n - 2\) is \(\frac{1}{6}\). What is the probability that the larger of the two integers chosen is greater than \(n - 2\)?
Total number of ways to choose \(2\) of \(n\) integers:
\(\frac{n(n - 1)}{2}\)
Number of ways to choose \(n - 2\) and an integer less than \(n - 2\):
There's one way to choose \(n - 2\), and there are \(n - 3\) ways to choose one of the integers less than \(n - 2\).
So, the number of ways to choose \(n - 2\) and an integer less than \(n - 2\) is \(1 × n - 3 = n - 3\).
Find Favorable/Total:
\(\frac{Favorable}{Total} = \frac{n - 3}{n(n - 1)/2} = \frac{1}{6}\)
Solve for \(n\):
\(12n - 36 = n^2 - n\)
\(n^2 - 13n + 36 = 0\)
\((n - 9)(n - 4) = 0\)
From the passage, \(n > 5\).
\(n = 9\)
Answer the question asked:
Total ways to choose \(2\) of \(9 = \frac{9 × 8}{2!} = 36\)
Number of ways to choose \(2\) of \(9\) integers such that the larger of the two integers chosen is greater than \(9 - 2\), meaning that the larger is \(9\) or \(8\):
\((1 × 8) + (1 × 7) = 15\)
Probability that the larger of the two integers chosen is greater than \(9 - 2\):
\(\frac{15}{36} = \frac{5}{12}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{5}{12}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{5}{6}\)
Correct answer: C
Total number of ways to choose \(2\) of \(n\) integers:
\(\frac{n(n - 1)}{2}\)
Number of ways to choose \(n - 2\) and an integer less than \(n - 2\):
There's one way to choose \(n - 2\), and there are \(n - 3\) ways to choose one of the integers less than \(n - 2\).
So, the number of ways to choose \(n - 2\) and an integer less than \(n - 2\) is \(1 × n - 3 = n - 3\).
Find Favorable/Total:
\(\frac{Favorable}{Total} = \frac{n - 3}{n(n - 1)/2} = \frac{1}{6}\)
Solve for \(n\):
\(12n - 36 = n^2 - n\)
\(n^2 - 13n + 36 = 0\)
\((n - 9)(n - 4) = 0\)
From the passage, \(n > 5\).
\(n = 9\)
Answer the question asked:
Total ways to choose \(2\) of \(9 = \frac{9 × 8}{2!} = 36\)
Number of ways to choose \(2\) of \(9\) integers such that the larger of the two integers chosen is greater than \(9 - 2\), meaning that the larger is \(9\) or \(8\):
\((1 × 8) + (1 × 7) = 15\)
Probability that the larger of the two integers chosen is greater than \(9 - 2\):
\(\frac{15}{36} = \frac{5}{12}\)
(A) \(\frac{1}{3}\)
(B) \(\frac{2}{5}\)
(C) \(\frac{5}{12}\)
(D) \(\frac{1}{2}\)
(E) \(\frac{5}{6}\)
Correct answer: C
