\[ 256 = 2^8 \]

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• The prime factorization of 48 is:
• \[ 48 = 2^4 \times 3^1 \]
• The prime factorization of 256 is:
• The GCD is found by taking the lowest power of all prime factors common to both numbers. Here, the only common prime factor is 2.
• For 2, the lowest power common to both factorizations is \(2^4\).
• Thus, the GCD of 48 and 256 is:

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$4 \times 3 \times 12 = 144$Therefore, the desired probability is:\frac{144}{256} = \frac{9}{16}Wait — but this counts only sequences with one pair and two singles, and one option unused. But is that the only case with exactly one repeated word? Yes, under the condition exactly two of the chosen options are the same and the other two are each different — this implies exactly one repetition, two others distinct. So the multiset is indeed {A,A,B,C} with B ≠ C ≠ A.But note: the problem says exactly two of the chosen options are the same — this could be misinterpreted as at least two are the same, but the phrasing exactly two suggests a single pair and the rest different — i.e., one word appears twice, the others appear once — which is standard in such puzzles.Thus, our count is correct.Hence, the probability is:But wait: $144 \div 16 = 9$, $256 \div 16 = 16$, yes: $\frac{144}{256} = \frac{9}{16}$?Wait: $144 / 16 = 9$, $256 / 16 = 16$, so yes, $\frac{144}{256} = \frac{9}{16}$? Actually:$144 = 16 \times 9$, $256 = 16 \times 16$, yes — so $\frac{144}{256} = \frac{9}{16}$? No: $16 \times 9 = 144$, $16 \times 16 = 256$, so $\frac{144}{256} = \frac{9}{16}$? Wait: $\frac{144}{256} = \frac{144 \div 16}{256 \div 16} = \frac{9}{16}$ — yes, correct.