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

  • question_answer
    In an entrance test there are multiple choice questions. There are four possible answers to each question of which one is correct. The probability that a student knows the answer to a question is 90%. If he gets the correct answer to a question, then the probability that he was guessing, is

    A)                 \[\frac{37}{40}\]  

    B)                 \[\frac{1}{37}\]

    C)                 \[\frac{36}{37}\]  

    D)                 \[\frac{1}{9}\]

    Correct Answer: B

    Solution :

               We define the following events :            \[{{A}_{1}}:\] He knows the answer.            \[{{A}_{2}}:\] He does not know the answer.            \[E:\] He gets the correct answer.            Then \[P({{A}_{1}})=\frac{9}{10},\,\,P({{A}_{2}})=1-\frac{9}{10}=\frac{1}{10},\,P\text{ }\left( \frac{E}{{{A}_{1}}} \right)=1,\]            \[P\left( \frac{E}{{{A}_{2}}} \right)=\frac{1}{4}\]            \[\therefore \] Required probability                 \[=P\left( \frac{{{A}_{2}}}{E} \right)=\frac{P({{A}_{2}})P(E/{{A}_{2}})}{P({{A}_{1}})P(E/{{A}_{1}})+P({{A}_{2}})P(E/{{A}_{2}})}=\frac{1}{37}.\]


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