If A and B are two events such that P(A | B) | vedclass

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(A) Given: $P(A | B) = 0.6$,$P(B | A) = 0.3$,and $P(A) = 0.1$.Using the definition of conditional probability,$P(B | A) = \frac{P(A \cap B)}{P(A)}$.Su

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$0.88$

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(A) Given: $P(A | B) = 0.6$,$P(B | A) = 0.3$,and $P(A) = 0.1$.Using the definition of conditional probability,$P(B | A) = \frac{P(A \cap B)}{P(A)}$.Substituting the values: $0.3 = \frac{P(A \cap B)}{0.1} \implies P(A \cap B) = 0.03$.Next,$P(A | B) = \frac{P(A \cap B)}{P(B)}$.Substituting the values: $0.6 = \frac{0.03}{P(B)} \implies P(B) = \frac{0.03}{0.6} = 0.05$.By De Morgan's Law,$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.Using the addition rule,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$P(A \cup B) = 0.1 + 0.05 - 0.03 = 0.12$.Therefore,$P(\bar{A} \cap \bar{B}) = 1 - 0.12 = 0.88$.

If $A$ and $B$ are two events such that $P(A | B) = 0.6$,$P(B | A) = 0.3$,and $P(A) = 0.1$,then $P(\bar{A} \cap \bar{B})$ equals:

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