If A and B are two events such that `P(A)!=0`and `P(B | A) = 1`; then(A) `AsubB` (B) `BsubA` (C) `B=varphi` (D) `A=varphi`A. `AsubB` but `A!=B`B. `BsubA`C. `B=phi`D. `A=phi`

If A and B are two events such that `P(A)!=0`and `P(B | A) = 1`; then(

1.	If A and B are two events such that `P(A)!=0`and `P(B \| A) = 1`; then(A) `AsubB` (B) `BsubA` (C) `B=varphi` (D) `A=varphi`A. `AsubB` but `A!=B`B. `BsubA`C. `B=phi`D. `A=phi`
Answer» Correct Answer - A (a) Given `P(A)!=0` and `P(B/A)=1impliesP(B/A)=(P(AnnB))/(P(A))=1` `impliesP(AnnB)=P(A)impliesAnnB=AimpliesAsubB`