A) A and B are equally likely
B) \[P\left( A\cap B' \right)=0\]
C) \[P\left( A'\cap B \right)=0\]
D) \[P(A)+P(B)=1\]
Correct Answer: D
Solution :
Let A and B be two events such that \[P(A\cup B)=P(A\cap B)\] and \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] option (a) : since \[P(A\cup B)=P(A\cap B)\] (given) therefore A and B are equally likely Suppose option (b) and option (c) are correct. \[\therefore \]\[P(A\cap B')=0\]and\[P('A\cap B)=0\] \[\Rightarrow \]\[P(A)-P(A\cap B)=0\]and\[P(B)=P(A\cap B)\] Thus \[P(A)=P(B)=P(A\cap B)=P(A\cup B)\] [\[\because \]Given \[P(A\cap B)=P(A\cup B)\]] Also, we know \[P(A\cup B)=P(A)+P(B)-P(A\cap B)\] \[=P(A\cap B)+P(A\cap B)-P(A\cap B)\] \[=P(A\cap B)\] which is true from given condition Hence, option (a), (b) and (c) are correct.You need to login to perform this action.
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