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

  • question_answer
    A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is

    A)                 \[\frac{3}{8}\]       

    B)                 \[\frac{1}{5}\]

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

    D)                 None of these

    Correct Answer: A

    Solution :

               Let \[E\] denote the event that a six occurs and \[A\] the event that the man reports that it is a ?6?, we have            \[P(E)=\frac{1}{6},\,\,P({E}')=\frac{5}{6},\,\,P(A/E)=\frac{3}{4}\] and \[P(A/{E}')=\frac{1}{4}\]            From Baye?s theorem,            \[P(E/A)=\frac{P(E).P(A/E)}{P(E).P(A/E)+P({E}').P(A/{E}')}\]                                 \[=\frac{\frac{1}{6}\times \frac{3}{4}}{\frac{1}{6}\times \frac{3}{4}+\frac{5}{6}\times \frac{1}{4}}=\frac{3}{8}.\]


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