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Let A, B, and C be three events. The probability that exactly one of A and B occurs is 1, that exactly one of B and C occurs is 1/2, and that exactly one of C and A occurs is 1/4. If the probability that all three events occur together is a², then the probability that at least one of A, B, or C occurs is

1. 1/2
2. < 1/2
3. > 1/2
4. None of these

Correct answer: > 1/2

Solution

Adding the three 'exactly one' probabilities: [P(A)+P(B)-2P(AB)]+... = 1 + 1/2 + 1/4 = 7/4, so (sum of singles) - (sum of pairwise) = 7/8. Then P(at least one) = (sum singles) - (sum pairs) + P(ABC) = 7/8 + a^2 >= 7/8 > 1/2.

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