We point out that A∩Bˉ=A∖(A∩B)A\cap \bar{B}=A\setminus( A\cap {B})A∩Bˉ=A∖(A∩B) . Therefore P(A∩Bˉ)=P(A)−P(A∩B)P(A\cap \bar{B})=P(A)-P(A\cap {B})P(A∩Bˉ)=P(A)−P(A∩B) .
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