If A and B are two events such that $P(A)=0.4$, $P(B)=0.8$ and $P(B/A)=0.6$, then $P(\overline{A} \cap B)$ is: (a) 0.56 (b) 0.96 (c) 0.66 (d) 0.24

If A and B are two events such that P(A)=0.4, P(B)=0.8 and P(B/A)=0.6, then P(overlineA ∩ B) is: (a) 0.56 (b)

Question 67 of 89

P(A)=0.4

(a) 0.56

(b) 0.96

(d) 0.24

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$P(\overline{A}\cap B) = P(B)-P(A\cap B) = 0.8-0.24 = 0.56$ .

First find $P(A\cap B)$ using the conditional probability formula: $P(B/A) = \dfrac{P(A\cap B)}{P(A)}$ , so $P(A\cap B) = P(B/A)\cdot P(A) = 0.6\times0.4 = 0.24$ .

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Q.If A and B are two events such that $P(A)=0.4$ , $P(B)=0.8$ and $P(B/A)=0.6$ , then $P(\overline{A} \cap B)$ is:

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