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

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(D) We know that $P(A \cup B) \ge \max\{P(A), P(B)\} = \frac{2}{3}$.Also,$P(A \cap B) \le \min\{P(A), P(B)\} = \frac{1}{2}$.Using the formula $P(A \ca

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(D) We know that $P(A \cup B) \ge \max\{P(A), P(B)\} = \frac{2}{3}$.Also,$P(A \cap B) \le \min\{P(A), P(B)\} = \frac{1}{2}$.Using the formula $P(A \cap B) = P(A) + P(B) - P(A \cup B)$,and since $P(A \cup B) \le 1$,we have $P(A \cap B) \ge P(A) + P(B) - 1 = \frac{1}{2} + \frac{2}{3} - 1 = \frac{1}{6}$.Thus,$\frac{1}{6} \le P(A \cap B) \le \frac{1}{2}$.For $P(A' \cap B)$,we use $P(A' \cap B) = P(B) - P(A \cap B)$.Substituting the bounds for $P(A \cap B)$,we get $\frac{2}{3} - \frac{1}{2} \le P(A' \cap B) \le \frac{2}{3} - \frac{1}{6}$.This simplifies to $\frac{1}{6} \le P(A' \cap B) \le \frac{1}{2}$.Since all statements are correct,the answer is $(d)$.

If $A$ and $B$ are two events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{2}{3},$ then

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