JEE Main & Advanced Mathematics Probability Question Bank Conditional probability bayes theorem

  • question_answer
    One ticket is selected at random from 100 tickets numbered 00, 01, 02, ...... 98, 99. If X and Y denote the sum and the product of the digits on the tickets, then \[P\,(X=9/Y=0)\] equals

    A)                 \[\frac{1}{19}\]    

    B)                 \[\frac{2}{19}\]

    C)                 \[\frac{3}{19}\]    

    D)                 None of these

    Correct Answer: B

    Solution :

               Event \[(Y=0)\] is \[\{00,\,\,01,\,\,09,\,\,10,\,\,20,\,\,..........90\}\]            Also \[(X=9)\cap (Y=0)=09,\,\,90,\] we have            \[P(Y=0)=\frac{19}{100}\] and \[P(X=9)\cap (Y=0)=\frac{2}{100}\]            Hence required probability                                 \[=P\left\{ (X=9)/(Y=0) \right\}=\frac{\left\{ P(X=9)\cap (Y=0) \right\}}{P(Y=0)}=\frac{2}{19}\].


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