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Exercise 12.3 · Q2

Q.If AA and BB are two events such that P(A∪B)=0.7P(A\cup B) = 0.7, P(A∩B)=0.2P(A\cap B) = 0.2, and P(B)=0.5P(B) = 0.5, then show that AA and BB are independent.

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✓ Free question

Step 1. Recover P(A)P(A). Using the Addition Theorem, P(A∪B)=P(A)+P(B)−P(A∩B)P(A\cup B)=P(A)+P(B)-P(A\cap B), so P(A)=P(A∪B)−P(B)+P(A∩B)=0.7−0.5+0.2=0.4P(A)=P(A\cup B)-P(B)+P(A\cap B)=0.7-0.5+0.2=0.4.

Step 2. Test the independence condition. Independence needs P(A∩B)=P(A)⋅P(B)P(A\cap B)=P(A)\cdot P(B). Compute the right-hand side: P(A)⋅P(B)=0.4×0.5=0.2P(A)\cdot P(B)=0.4\times0.5=0.2.

Step 3. Compare with the given value. The given P(A∩B)=0.2P(A\cap B)=0.2 exactly matches P(A)⋅P(B)=0.2P(A)\cdot P(B)=0.2.

✓Final answer

P(A)=0.4P(A)=0.4, and since P(A∩B)=0.2=P(A)⋅P(B)P(A\cap B)=0.2=P(A)\cdot P(B), the events AA and BB are independent.

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