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JEE Main & Advanced AIEEE Solved Paper-2008

  • question_answer
    It is given that the events A and B are such that \[P\left( A \right)=\frac{1}{4},\,P\left( A|B \right)=\frac{1}{2}\] and \[\,P\left( B|A \right)=\frac{2}{3}\]. Then \[P\left( B \right)\] is       AIEEE  Solved  Paper-2007

    A) \[\frac{2}{3}\]

    B)                                        \[\frac{1}{2}\]

    C) \[\frac{1}{6}\]                  

    D)        \[\frac{1}{3}\]

    Correct Answer: D

    Solution :

                    \[P\left( A \right)=\frac{1}{4}\] \[P\left( \frac{A}{B} \right)=\frac{P\left( A\cap B \right)}{P\left( B \right)}\Rightarrow \frac{1}{2}=\frac{P\left( A\cap B \right)}{P\left( B \right)}\]           ? (i) \[P\left( \frac{B}{A} \right)=\frac{P\left( A\cap B \right)}{P\left( A \right)}\Rightarrow \frac{2}{3}=\frac{P\left( A\cap B \right)}{1/4}\]\[\Rightarrow P\left( A\cap B \right)=\frac{1}{6}\]. Putting the value of \[P\left( A\cap B \right)\] in (i) \[\Rightarrow P\left( B \right)=2\times \frac{1}{6}=\frac{1}{3}\]


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