A) Not independent
B) Also independent
C) Mutually exclusive
D) None of these
Correct Answer: B
Solution :
Since \[A\cap \bar{B}\] and \[A\cap B\] are mutually exclusive events such that \[A=(A\cap \bar{B})\cup (A\cap B)\] \ \[P(A)=P(A\cap \bar{B})+P(A\cap B)\] \[\Rightarrow P(A\cap \bar{B})=P(A)-P(A\cap B)\]\[=P(A)-P(A)P(B)\] \[(\because \,\,\,A,\,B\] are independent) \[\Rightarrow P(A\cap \bar{B})\]\[=P(A)(1-P(B))=P(A)P(\bar{B})\] \[\therefore \,\,\,A\] and \[\bar{B}\] are also independent.You need to login to perform this action.
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