P(A cup B) = P(A cap B) if and only if the re | vedclass

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(C) We know that the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Given the condition $P(A \cup B) = P(A \cap

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$P(A) = P(B)$

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(C) We know that the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Given the condition $P(A \cup B) = P(A \cap B)$,we substitute this into the formula:$P(A \cap B) = P(A) + P(B) - P(A \cap B)$.Rearranging the terms,we get $2P(A \cap B) = P(A) + P(B)$.However,for the specific case where $P(A \cup B) = P(A \cap B)$,it implies that $P(A \setminus B) = 0$ and $P(B \setminus A) = 0$,which means $P(A) = P(A \cap B) = P(B)$.Therefore,the correct relation is $P(A) = P(B)$.

$P(A \cup B) = P(A \cap B)$ if and only if the relation between $P(A)$ and $P(B)$ is

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