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If P (A) = 0.4, P (B) = 0.8, P (B/A) = 0.6. Find P (A/B) and P (A ∪ B). |
Answer» We have, P (A ∪ B) = P (A)+ P (B)- P (A ∩ B) From equation, \(P(\cfrac{B}{A})=\cfrac{P(A\cap B)}{P(A)}\) P (B ∩ A) = \(P(\cfrac{B}{A})\) x P(A) = 0.6 x 0.4 = 0.24 Hence, \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}=\cfrac{0.24}{0.8}\) = 0.3 And P (A ∪ B) = P (A)+ P (B)- P (A ∩ B) = 0.4 + 0.8 - 0.24 = 0.96 |
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