A) \[P(A)+P(B)-P(A)P(B)\]
B) \[P(A)-P(B)\]
C) \[P(A)+P(B)\]
D) \[P(A)+P(B)+P(A)P(B)\]
Correct Answer: A
Solution :
Key Idea: If \[A\] and \[B\] are independent, then \[P(A\cap B)=P(A)P(B)\] \[\therefore \] \[P(A\cup B)=P(A+B)\] \[=P(A)+P(B)-P(A\cap B)\] \[=P(A)+P(B)-P(A)P(B)\] Note: If \[A\] and \[B\] are two mutually exclusive events, then\[P(A\cap B)=0\].You need to login to perform this action.
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