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Questions  

 Let 0<P(A)<1,0<P(B)<1 and P(AB)=P(A)+P(B)P(A)P(B), then 

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a
P(A∣B)=0
b
P(B∣A)=0
c
PA′∩B′=PA′PB′
d
P(A∣B)+P(B∣A)=1

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detailed solution

Correct option is C

We know that P(A∪B)=P(A)+P(B)−P(A∩B)⇒P(A)+P(B)−P(A)P(B)=P(A)+P(B)−P(A∩B)⇒P(A)⋅P(B)=P(A∩B)⇒A and B are independent.


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