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It is given that the events A\; \; and\; \; B are such that P(A)=\frac{1}{4},\; P(A\mid B)=\frac{1}{2}\; \; and\; \; P(B\mid A)=\frac{2}{3}.  Then P(B) is

  • Option 1)

    \frac{1}{2}\;

  • Option 2)

    \; \frac{1}{6}\;

  • Option 3)

    \; \frac{1}{3}\;

  • Option 4)

    \; \frac{2}{3}

 

Answers (1)

best_answer

As we learnt in 

Conditional Probability -

 

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

and

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

 

- wherein

where P\left ( \frac{A}{B} \right ) probability of A when B already happened.

 

 

P\left ( B|A \right ) P\left ( A \right ) = P \left ( A|B \right ) P\left ( B \right )=P\left ( A\cap B \right )

\Rightarrow \frac{1}{4}\times \frac{2}{3} =\frac{1}{2}\times P\left ( B \right )

\Rightarrow P\left ( B \right ) = \frac{1}{3}


Option 1)

\frac{1}{2}\;

Incorrect

Option 2)

\; \frac{1}{6}\;

Incorrect

Option 3)

\; \frac{1}{3}\;

Correct

Option 4)

\; \frac{2}{3}

Incorrect

Posted by

prateek

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