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Question
If 0 < P(A) < 1, 0 < P(B) < 1 and P(A ∪ B) = P(A) + P(B) – P(A)P(B), then
Options
P(B/A) = P(B) – P(A)
P[(A ∪ B)c] = P(Ac) + P(Bc)
P[(A ∪ B)c] = P(Ac) + P(Bc)
P[(Ac ∪ B)c] = P(Ac) + P(Bc)
MCQ
Solution
P[(A ∪ B)c] = P(Ac) + P(Bc)
Explanation:
Given P(A ∪ B) = P(A) + P(B) – P(A).P(B)
⇒ 1 – P[(A ∪ B)c] = P(A) + P(B) [1 – P(A)]
⇒ 1 – P[(A ∪ B)c] = P(A) + P(B) P(Ac)
⇒ 1 – P(A) – P(B) P(Ac) = P[(A ∪ B)c]
⇒ P(Ac) – P(B) P(Ac) = P[(A ∪ B)c]
⇒ P(Ac) [1 – P(B)] = P[(A ∪ B)c]
⇒ P(Ac) . P(Bc) = P[(A ∪ B)c]
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