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2019 AMC 12A Problem 20

Problem 20 of 25HarderProbability & StatisticsProblem-Solving Techniques

Real numbers between 00 and 1,1, 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| \gt \dfrac{1}{2}?

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Solution

Each variable equals 00 with probability 14,\tfrac14, equals 11 with probability 14,\tfrac14, and is uniform on [0,1][0, 1] with probability 12.\tfrac12. Considering the nine combinations of types: the pairs (0,1)(0, 1) and (1,0)(1, 0) each contribute 116.\tfrac{1}{16}. Each of the four point-versus-uniform cases contributes 116.\tfrac{1}{16}. The uniform-versus-uniform case contributes 14⋅14=116.\tfrac14 \cdot \tfrac14 = \tfrac{1}{16}. The total is 2+4+116=716.\dfrac{2 + 4 + 1}{16} = \dfrac{7}{16}. Thus, the correct answer is B.
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