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So the probability is $ \frac916 $. - AMAZONAWS
So the probability is $ \frac916 $. - AMAZONAWS
📅 March 6, 2026
👤 scraface
Mar 06, 2026
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📌 We are to compute the probability that in 4 independent choices among 4 options (say labeled A, B, C, D, with equal likelihood), exactly one option appears twice, and two other distinct options appear once each, with the fourth position filled by a fourth distinct option — but wait: this would require 4 distinct options total, and one repeated. Since only 4 positions exist, and we want exactly one option repeated, and the other two being different, the only valid pattern is: one option appears twice, two others appear once, and the fourth option appears zero times — but that’s only 4 choices total. So the pattern is: one option repeated twice, and two other distinct options appearing once each — that uses up 2 + 1 + 1 = 4 choices, with one option appearing twice and two others once each, and one option not used at all. So the multiset of choices is of the form {A, A, B, C}, where A, B, C are distinct, and D is unused.
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