If P(A)=x,P(B)=yP(A) = x, P(B) = yP(A)=x,P(B)=y, and P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)P(A∪B)=P(A)+P(B), then:
A,BA, BA,B are independent.
A∩B=∅A \cap B = \emptysetA∩B=∅.
P(A∩B)=xyP(A \cap B) = xyP(A∩B)=xy.
P(A∣B)=P(A)P(A|B) = P(A)P(A∣B)=P(A).