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Question: If A and B are two events such that \(P ( A \cup B ) + P ( A \cap B ) = \frac { 7 } { 8 }\) and ...

If A and B are two events such that

P(A∪B)+P(A∩B)=78P ( A \cup B ) + P ( A \cap B ) = \frac { 7 } { 8 } and P(A)=2P(B)P ( A ) = 2 P ( B ) then P(A)=P ( A ) =

A

712\frac { 7 } { 12 }

B

724\frac { 7 } { 24 }

C

512\frac { 5 } { 12 }

D

1724\frac { 17 } { 24 }

Answer

712\frac { 7 } { 12 }

Explanation

Solution

Since we have

P(A∪B)+P(A∩B)=P(A)+P(B)P ( A \cup B ) + P ( A \cap B ) = P ( A ) + P ( B ) =P(A)+P(A)2= P ( A ) + \frac { P ( A ) } { 2 }