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JEE Main & Advanced Mathematics Probability Question Bank Self Evaluation Test - Probability-II

  • question_answer
    For a biased dice, the probability for the different faces to turn up are
    Face 1 2 3 4 5 6
    P 0.01 0.32 0.21 0.15 0.05 0.147
    The dice is tossed and it is told that either the face 1 or face 2 has shown up, then the probability that it is face, 1, is

    A) \[\frac{16}{21}\]

    B) \[\frac{1}{10}\]

    C) \[\frac{5}{16}\]

    D) \[\frac{5}{21}\]

    Correct Answer: D

    Solution :

    [d] Let E: ?face 1 comes up? and F: ?face 1 or 2 comes up? \[\Rightarrow E\cap F=E\]                        \[(\therefore E\subset F)\] \[\therefore P(E)=0.10\] and \[P(F)=P(1)+P(2)=0.10+0.32=0.42\] Hence, required probability \[=P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{P(E)}{P(F)}=\frac{0.10}{0.42}=\frac{5}{21}\]


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