If P(A) = 0.45, P(B) = 0.20, P(A⋂B) = 0.15, then =

Engineering

Mathematics

Conditional Probability

Question

If P(A) = 0.45, P(B) = 0.20, P(A⋂B) = 0.15, then $P (\bar{B} \cap \bar{B})$ =

0.38

0.39

0.40

0.50

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P(A) = 0.45, P(B) = 0.20

P(A∩B) = 0.15

⇒ P(AUB) = P(A) + P(B) – P(A∩B)

⇒ P(AUB) = 0.45 + 0.20 – 0.15

⇒ P(AUB) = 0.50

⇒ $P (\bar{A} \cap \bar{B}) = 1 - P (A \cup B)$

⇒ $P (\bar{A} \cap \bar{B}) = 1 - 0.50$

⇒ $P (\bar{A} \cap \bar{B}) = 0.50$

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