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AMC 12 2019a Problem 20

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Real numbers between 00 and 11, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 00 if the second flip is heads and 11 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval [0,1][0,1]. Two random numbers xx and yy are chosen independently in this manner. What is the probability that ∣x−y∣>12|x-y| > \tfrac{1}{2}?

(A) 13(B) 716(C) 12(D) 916(E) 23\textbf{(A) } \frac{1}{3} \textbf{(B) } \frac{7}{16} \textbf{(C) } \frac{1}{2} \textbf{(D) } \frac{9}{16} \textbf{(E) } \frac{2}{3}

Dataset text AMC 12 2000 to 2025. Quoted with source.

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