But then if P(A)=0, P(B|A) + P(¬B|A) = 2, which seems undesirable.
Since P(A|B) is properly undefined, any attempt to define it will have some negative features.
But then if P(A)=0, P(B|A) + P(¬B|A) = 2, which seems undesirable.
Since P(A|B) is properly undefined, any attempt to define it will have some negative features.