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JEE Main & Advanced Mathematics Probability Question Bank Odds in favour and odds against Addition theorem on probability

  • question_answer
    Let A and B be two events such that \[P\overline{(A\cup B)}=\frac{1}{6},P(A\cap B)=\frac{1}{4}\] and \[P(\bar{A})=\frac{1}{4},\] where \[\bar{A}\] stands for complement of event A. Then events A and B are       [AIEEE 2005]

    A)                 Independent but not equally likely

    B)                 Mutually exclusive and independent

    C)                 Equally likely and mutually exclusive

    D)                 Equally likely but not independent

    Correct Answer: A

    Solution :

                       \[P\left( \overline{A\cup B} \right)=\frac{1}{6};\,\,P\left( A\cap B \right)=\frac{1}{4}\],                    \[P\left( {\bar{A}} \right)=\frac{1}{4}\Rightarrow P\left( A \right)=\frac{3}{4}\],                    \[P\left( \overline{A\cup B} \right)=1-P\left( A\cup B \right)=1-P\left( A \right)-P\left( B \right)+P\left( A\cap B \right)\]                                 Þ \[\frac{1}{6}=\frac{1}{4}-P\left( B \right)+\frac{1}{4}\]Þ\[P\left( B \right)=\frac{1}{3}\].                    Since \[P\left( A\cap B \right)=P\left( A \right)P\left( B \right)\] and \[P\left( A \right)\ne P\left( B \right)\]                                 \ A and B are independent but not equally likely.


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