If P(A) = frac{6}{11},P(B) = frac{5}{11},and | vedclass

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Question

(A) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given values: $\frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P(A \cap B)$.$

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(A) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given values: $\frac{7}{11} = \frac{6}{11} + \frac{5}{11} - P(A \cap B)$.$\frac{7}{11} = \frac{11}{11} - P(A \cap B)$.$P(A \cap B) = 1 - \frac{7}{11} = \frac{4}{11}$.Now,the conditional probability is given by $P(B | A) = \frac{P(A \cap B)}{P(A)}$.$P(B | A) = \frac{\frac{4}{11}}{\frac{6}{11}} = \frac{4}{6} = \frac{2}{3}$.

If $P(A) = \frac{6}{11}$,$P(B) = \frac{5}{11}$,and $P(A \cup B) = \frac{7}{11}$,find $P(B | A)$.

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