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

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(D) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Therefore, $P(A) + P(B) = P(A \cup B) + P(A \cap B)$.Given that $\frac{3}{4} \leq P(A \cup

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Both $A$ and $C$ are correct

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(D) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Therefore, $P(A) + P(B) = P(A \cup B) + P(A \cap B)$.Given that $\frac{3}{4} \leq P(A \cup B) \leq 1$ and $\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}$.Adding these inequalities, we get:$\frac{3}{4} + \frac{1}{8} \leq P(A \cup B) + P(A \cap B) \leq 1 + \frac{3}{8}$.$\frac{6+1}{8} \leq P(A) + P(B) \leq \frac{8+3}{8}$.$\frac{7}{8} \leq P(A) + P(B) \leq \frac{11}{8}$.Thus, both $P(A) + P(B) \geq \frac{7}{8}$ and $P(A) + P(B) \leq \frac{11}{8}$ are correct.

If $A$ and $B$ are two events such that $P(A \cup B) \geq \frac{3}{4}$ and $\frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8}$, then which of the following is true?

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