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

  • question_answer
    For a biased die, the probabilities for different faces to turn up are
    Face : 1 2 3 4 5 6
    Probability : 0.2 0.22 0.11 0.25 0.05 0.17
    The die is tossed and you are told that either face 4 or face 5 has turned up. The probability that it is face 4 is

    A)                 \[\frac{1}{6}\]       

    B)                 \[\frac{1}{4}\]

    C)                 \[\frac{5}{6}\]       

    D)                 None of these

    Correct Answer: C

    Solution :

                      Let \[A\] be the event that face 4 turns up and \[B\] be the event that face 5 turns up then \[P(A)=0.25,\] \[P(B)=0.05\]. Since \[A\] and \[B\] are mutually exclusive, so \[P(A\cup B)=P(A)+P(B)=0.25+0.05=0.30\].            We have to find \[P\left( \frac{A}{A\cup B} \right),\] which is equal to                 \[P\frac{[A\cap (A\cup B)]}{P(A\cup B)}=\frac{P(A)}{P(A\cup B)}=\frac{0.25}{0.30}=\frac{5}{6}\].


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