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JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    Three numbers are chosen at random without replacement from {1, 2, 3, ..., 8}. The probability that their minimum is 3, given that their maximum is 6, is :   AIEEE  Solved  Paper-2012

    A) \[\frac{3}{8}\]                                      

    B) \[\frac{1}{5}\]

    C) \[\frac{1}{4}\]                                      

    D) \[\frac{2}{5}\]

    Correct Answer: B

    Solution :

                 Let Event (Given :               {1, 2, 3,.........8}) A : Maximum of three numbers is 6. B : Minimum of three numbers is 3 \[P\left( \frac{B}{A} \right)=\frac{P(B\cap A)}{P(A)}=\frac{^{2}{{C}_{1}}}{^{5}{{C}_{2}}}=\frac{2}{10}=\frac{1}{5}\]


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