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The likelihoods of three events X, Y, and Z are P(X) = 0.6, P(Y) = 0.4, and P(Z) = 0.5. If P(X ∪ Y) = 0.8, P(X ∩ Z) = 0.3, P(X ∩ Y ∩ Z) = 0.2, and P(X ∪ Y ∪ Z) ≥ 0.85, then which of the following is true?

1. P(X ∩ Y) ≤ 0.35
2. P(Y ∩ Z) ≥ 0.2
3. P(Y ∩ Z) ≤ 0.35
4. P(X ∩ Y) > 0.25

Correct answer: P(X ∩ Y) ≤ 0.35

Solution

Using the principle of inclusion-exclusion, we can derive the relationship between the probabilities of the events X, Y, and Z, and their intersections, which ultimately leads to the conclusion that P(X ∩ Y) ≤ 0.35.

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