If A and B are two events such that P(A∪B) = P(A∩B), then the incorrect statement amongst the following statements is:

P(A∪B) = P(A) + P(B) - P(A∩B)
Given that P(A∪B) = P(A∩B),
Then P(A) + P(B) - P(A∩B) = P(A∩B)
P(A) + P(B) = 2P(A∩B)
If A and B are mutually exclusive events, then P(A∩B) = 0
Then P(A) + P(B) = 0 which is not possible.
If A and B are independent events, then P(A∩B) = P(A)P(B)
Then P(A) + P(B) = 2P(A)P(B)
If P(A) = P(B) = 1, then 2 = 2, so this is possible.
If P(A) = P(B) = 0, then 0 = 0, so this is possible.
If P(A) = 1 and P(B) = 0, then 1 = 0, which is not possible.
If P(A) = 0 and P(B) = 1, then 1 = 0, which is not possible.
If A and B are equally likely, P(A) = P(B)
Then 2P(A) = 2P(A∩B) which implies P(A) = P(A∩B)
If P(A∩B') = 0, then A∩B' is an empty set, then A is contained in B.
If P(A'∩B) = 0, then A'∩B is an empty set, then B is contained in A.
If A ⊂ B, then P(A) ≤ P(B) and P(A∪B) = P(B)
Also P(A∩B) = P(A)
Then P(B) = P(A), so A and B are equally likely
If P(A) + P(B) = 1, then it is possible if P(A) = P(B) = 0.5
Then P(A∪B) = 1 and P(A∩B) = 0.5, which is incorrect
Thus, P(A) + P(B) = 1 is incorrect