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J & K CET Engineering J and K - CET Engineering Solved Paper-2004

  • question_answer
    In a test, an examines either guesses or copies or knows the answer to a multiple choice questions with four choices. The probability that he makes a guess; is \[\frac{1}{3}\] and the probability that he copies the answer is \[\frac{1}{6}\].The probability that his answer is correct given that he copied it is\[\frac{1}{8}\]. The probability that his answer is correct, given that he guessed it is\[\frac{1}{4}\]. The probability that they knew the answer to the questions given that he correctly answered is

    A) \[\frac{24}{31}\]

    B) \[\frac{31}{24}\]

    C) \[\frac{24}{29}\]

    D) \[\frac{29}{24}\]

    Correct Answer: C

    Solution :

    Let \[{{E}_{1}}\] = event that the examines guesses the answer \[{{E}_{2}}\] = event that he copies the answer \[{{E}_{3}}\] = event that he knows the answer and E = event that he answers correctly. Then, \[P({{E}_{1}})=\frac{1}{3},\,\,P({{E}_{2}})=\frac{1}{6}\] and  \[P({{E}_{3}}),\,\,=1-\left( \frac{1}{3}+\frac{1}{6} \right)=\frac{1}{2}\] \[\therefore \]  Required probability \[=P\left( \frac{{{E}_{3}}}{E} \right)\] \[=\frac{P(E/{{E}_{3}}).P({{E}_{3}})}{P(E/{{E}_{1}}).P(E/{{E}_{2}}).P({{E}_{2}})+P(E/{{E}_{3}}).P({{E}_{3}})}\] \[=\frac{1\times \frac{1}{2}}{\left( \frac{1}{4}\times \frac{1}{3} \right)+\left( \frac{1}{8}\times \frac{1}{6} \right)+\left( 1\times \frac{1}{2} \right)}=\frac{24}{29}\]


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