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JEE Main & Advanced Mathematics Probability Question Bank Self Evaluation Test - Probability-II

  • question_answer
    For k=1, 2, 3 the box \[{{B}_{k}}\] contains k red balls and \[(k+1)\] white balls. Let \[P({{B}_{1}})=\frac{1}{2},P({{B}_{2}})=\frac{1}{3}\]and \[P({{B}_{3}})=\frac{1}{6}.\] A box is selected at random and a ball is drawn from it, if a red ball is drawn, then the probability that it has come from box \[{{B}_{2}}\], is

    A) \[\frac{35}{78}\]

    B) \[\frac{14}{39}\]

    C) \[\frac{10}{13}\]

    D) \[\frac{12}{13}\]

    Correct Answer: B

    Solution :

    [b] In a box, \[{{B}_{1}}=1R,2W;{{B}_{2}}=2R,3W\] and \[{{B}_{3}}=3R,4W\] Also, given that, \[P({{B}_{1}})=\frac{1}{2},P({{B}_{2}})=\frac{1}{3}\] and \[P({{B}_{3}})=\frac{1}{6}\] \[\therefore P\left( \frac{{{B}_{2}}}{R} \right)\] \[=\frac{P({{B}_{2}})P\left( \frac{R}{{{B}_{2}}} \right)}{P({{B}_{1}})P\left( \frac{R}{{{B}_{1}}} \right)+P({{B}_{2}})P\left( \frac{R}{{{B}_{2}}} \right)+P({{B}_{3}})P\left( \frac{R}{{{B}_{3}}} \right)}\] \[=\frac{\frac{1}{3}\times \frac{2}{5}}{\frac{1}{2}\times \frac{1}{3}+\frac{1}{3}\times \frac{2}{5}+\frac{1}{6}\times \frac{3}{7}}=\frac{\frac{2}{15}}{\frac{1}{6}+\frac{2}{15}+\frac{1}{14}}=\frac{14}{39}.\]


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