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

  • question_answer
    A box contains N cons, m of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed is \[\frac{1}{2},\] while it is \[\frac{2}{3}\] when a biased coin is tossed. A coin is drawn from the box at random and it tossed twice. Then the probability that the coin drawn is fair, is

    A) \[\frac{9m}{8N+m}\]

    B) \[\frac{9m}{8N-m}\]

    C) \[\frac{9m}{8m-N}\]

    D) \[\frac{9m}{8m+N}\]  

    Correct Answer: A

    Solution :

    [a] \[{{E}_{1}}:\] Coin is fair, \[{{E}_{2}}:\] coin is biased, a second toss shows tail. \[P({{E}_{1}}/A)=\frac{P(A/{{E}_{1}})P({{E}_{1}})}{P(A/{{E}_{1}})P({{E}_{1}})+P(A/{{E}_{1}})P({{E}_{2}})}\] \[=\frac{\frac{m}{N}.\frac{1}{2}.\frac{1}{2}}{\frac{m}{N}.\frac{1}{2}.\frac{1}{2}+\frac{N-m}{N}.\frac{2}{3}.\frac{1}{3}}=\frac{9m}{8N+m}\]


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