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

  • question_answer
    There are 3 bags which are known to contain 2 white and 3 black balls; 4 white and 1 black balls and 3 white and 7 black balls respectively. A ball is drawn at random from one of the bags and found to be a black ball. Then the probability that it was drawn from the bag containing the most black balls is

    A)                 \[\frac{7}{15}\]    

    B)                 \[\frac{5}{19}\]

    C)                 \[\frac{3}{4}\]       

    D)                 None of these

    Correct Answer: A

    Solution :

               Consider the following events :            \[A\to \] Ball drawn is black; \[{{E}_{1}}\to \] Bag I is chosen;            \[{{E}_{2}}\to \] Bag II is chosen and \[{{E}_{3}}\to \] Bag III is chosen.            Then \[P({{E}_{1}})=({{E}_{2}})=P({{E}_{3}})=\frac{1}{3},\,\,P\left( \frac{A}{{{E}_{1}}} \right)=\frac{3}{5}.\]            \[P\left( \frac{A}{{{E}_{2}}} \right)=\frac{1}{5},\,\,P\left( \frac{A}{{{E}_{3}}} \right)=\frac{7}{10}\]            Required probability \[=P\left( \frac{{{E}_{3}}}{A} \right)\]                 \[=\frac{P({{E}_{3}})P(A/{{E}_{3}})}{P({{E}_{1}})P(A/{{E}_{1}})+P({{E}_{2}})P(A/{{E}_{2}})+P({{E}_{3}})P(A/{{E}_{3}})}=\frac{7}{15}\].


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