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Question: For two events A and B, if \(P ( A ) = P \left( \frac { A } { B } \right) = \frac { 1 } { 4 }\) an...

For two events A and B, if P(A)=P(AB)=14P ( A ) = P \left( \frac { A } { B } \right) = \frac { 1 } { 4 } and P(BA)=12P \left( \frac { B } { A } \right) = \frac { 1 } { 2 }, then

A

A and B are independent

B

P(A′B)=34P \left( \frac { A ^ { \prime } } { B } \right) = \frac { 3 } { 4 }

C

P(B′A′)=12P \left( \frac { B ^ { \prime } } { A ^ { \prime } } \right) = \frac { 1 } { 2 }

D

All of these

Answer

All of these

Explanation

Solution

P(BA)=12P \left( \frac { B } { A } \right) = \frac { 1 } { 2 } ⇒P(B∩A)P(A)=12\Rightarrow \frac { P ( B \cap A ) } { P ( A ) } = \frac { 1 } { 2 } ⇒P(B∩A)=18\Rightarrow P ( B \cap A ) = \frac { 1 } { 8 }

P(AB)=14P \left( \frac { A } { B } \right) = \frac { 1 } { 4 } ⇒P(A∩B)P(B)=14\Rightarrow \frac { P ( A \cap B ) } { P ( B ) } = \frac { 1 } { 4 } ⇒P(B)=12\Rightarrow P ( B ) = \frac { 1 } { 2 }

P(A∩B)=18=P(A)⋅P(B)P ( A \cap B ) = \frac { 1 } { 8 } = P ( A ) \cdot P ( B )

∴\therefore Events A and B are independent.

Now, P(A′B)=P(A′∩B)P(B)=P(A′)P(B)P(B)=34P \left( \frac { A ^ { \prime } } { B } \right) = \frac { P \left( A ^ { \prime } \cap B \right) } { P ( B ) } = \frac { P \left( A ^ { \prime } \right) P ( B ) } { P ( B ) } = \frac { 3 } { 4 }

and P(B′A′)=P(B′∩A′)P(A′)=P(B′)P(A′)P(A′)=12P \left( \frac { B ^ { \prime } } { A ^ { \prime } } \right) = \frac { P \left( B ^ { \prime } \cap A ^ { \prime } \right) } { P \left( A ^ { \prime } \right) } = \frac { P \left( B ^ { \prime } \right) P \left( A ^ { \prime } \right) } { P \left( A ^ { \prime } \right) } = \frac { 1 } { 2 }.