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Question: If *A* and *B* are two events such that \(P ( A \cup B ) = P ( A \cap B )\) , then the true relatio...

If A and B are two events such that P(A∪B)=P(A∩B)P ( A \cup B ) = P ( A \cap B ) , then the true relation is

A

P(A)+P(B)=0P ( A ) + P ( B ) = 0

B

P(A)+P(B)=P(A)P(BA)P ( A ) + P ( B ) = P ( A ) P \left( \frac { B } { A } \right)

C

P(A)+P(B)=2P(A)P(BA)P ( A ) + P ( B ) = 2 P ( A ) P \left( \frac { B } { A } \right)

D

None of these

Answer

P(A)+P(B)=2P(A)P(BA)P ( A ) + P ( B ) = 2 P ( A ) P \left( \frac { B } { A } \right)

Explanation

Solution

P(A∪B)=P(A)+P(B)−P(A∩B)P ( A \cup B ) = P ( A ) + P ( B ) - P ( A \cap B )

⇒P(A∩B)=P(A)+P(B)−P(A∩B)P ( A \cap B ) = P ( A ) + P ( B ) - P ( A \cap B ) {∵P(A∩B)=P(A∪B)}\{ \because P ( A \cap B ) = P ( A \cup B ) \}⇒ 2P(A∩B)=P(A)+P(B)⇒2P(A)⋅P(A∩B)P(A)2 \mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) = \mathrm { P } ( \mathrm { A } ) + \mathrm { P } ( \mathrm { B } ) \Rightarrow 2 \mathrm { P } ( \mathrm { A } ) \cdot \frac { \mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) } { \mathrm { P } ( \mathrm { A } ) } =P(A)+P(B)⇒2P(A)P(BA)=P(A)+P(B)= \mathrm { P } ( \mathrm { A } ) + \mathrm { P } ( \mathrm { B } ) \Rightarrow 2 \mathrm { P } ( \mathrm { A } ) \mathrm { P } \left( \frac { \mathrm { B } } { \mathrm { A } } \right) = \mathrm { P } ( \mathrm { A } ) + \mathrm { P } ( \mathrm { B } )