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Q.If P(A)=0.5P(A) = 0.5, P(B)=0.3P(B) = 0.3 (given that AA and BB are mutually exclusive), then P(A′∩B′)=P(A' \cap B') =

(1) 0.60.6
(2) 0.50.5
(3) 0.70.7
(4) 0.20.2
Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2026MCQ· 1mImportance★★★★★
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P(A∪B)=0.8P(A\cup B)=0.8, so P(A′∩B′)=1−0.8=0.2P(A'\cap B')=1-0.8=0.2.

Since A,BA,B are mutually exclusive, P(A∩B)=0P(A\cap B)=0, hence

P(A∪B)=P(A)+P(B)=0.5+0.3=0.8.P(A\cup B)=P(A)+P(B)=0.5+0.3=0.8.

By De Morgan's law A′∩B′=(A∪B)′A'\cap B'=(A\cup B)', so …

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