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1.1-6. If P (A) = 0.4, P (B) = 0.5, and P (A ∩ B) = 0.3, find(a) P (A ∩ B)(b) P (A ∪ B′)(c) P (A′ ∪ B′)

Question

1.1-6. If P (A) = 0.4, P (B) = 0.5, and P (A ∩ B) = 0.3, find

(a) P (A ∩ B)
(b) P (A ∪ B′)
(c) P (A′ ∪ B′)

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Solution

Sure, here are the solutions:

(a) P(A ∩ B) = 0.3. This is given in the problem.

(b) P(A ∪ B') = P(A) + P(B') - P(A ∩ B') First, we need to find P(B'), which is the probability of not B. Since P(B) = 0.5, P(B') = 1 - P(B) = 1 - 0.5 = 0.5. Next, we need to find P(A ∩ B'), which is the probability of A and not B. Since A and B are not mutually exclusive, P(A ∩ B') = P(A) - P(A ∩ B) = 0.4 -

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