For two given events A and B,P(A cap B) is: | vedclass

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(D) The addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Rearranging this gives $P(A \cap B) = P(A) + P(B) - P(A

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(D) The addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Rearranging this gives $P(A \cap B) = P(A) + P(B) - P(A \cup B)$,which matches option $C$.Since $0 \leq P(A \cup B) \leq 1$,we have $P(A \cup B) \leq 1 \implies P(A) + P(B) - P(A \cap B) \leq 1 \implies P(A \cap B) \geq P(A) + P(B) - 1$,which matches option $A$.Also,since $P(A \cup B) \geq 0$,we have $P(A) + P(B) - P(A \cap B) \geq 0 \implies P(A \cap B) \leq P(A) + P(B)$,which matches option $B$.Since options $A$,$B$,and $C$ are all mathematically correct,the correct choice is $D$.

For two given events $A$ and $B$,$P(A \cap B)$ is:

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