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

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(D) Given $P(A \cup B) = P(A \cap B)$.We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given condition,we get $P(A \cap B) = P(

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$P(A) + P(B) = 1$

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(D) Given $P(A \cup B) = P(A \cap B)$.We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given condition,we get $P(A \cap B) = P(A) + P(B) - P(A \cap B)$,which implies $P(A) + P(B) = 2P(A \cap B)$.Also,we know $P(A \cap B') = P(A) - P(A \cap B)$ and $P(A' \cap B) = P(B) - P(A \cap B)$.Since $P(A \cap B) \le P(A)$ and $P(A \cap B) \le P(B)$,and $P(A \cup B) = P(A \cap B)$ implies that $A \subseteq B$ and $B \subseteq A$ (i.e.,$A = B$),we have $P(A) = P(B) = P(A \cap B) = P(A \cup B)$.Thus,$P(A \cap B') = P(A) - P(A) = 0$ and $P(A' \cap B) = P(B) - P(B) = 0$.Also,$P(A) = P(B)$ implies $A$ and $B$ are equally likely.However,$P(A) + P(B) = 1$ is not necessarily true,as $P(A) + P(B) = 2P(A \cap B)$,which is only $1$ if $P(A \cap B) = 0.5$.

If $A$ and $B$ are two events such that $P(A \cup B) = P(A \cap B)$,then the incorrect statement amongst the following statements is

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