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Thread: Binomial distribution

  1. #1
    Member Vinod's Avatar
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    Post Binomial distribution

    This problem have been taken from the old paper of MASTER of COMMERCE Examination of well-known University in India.
    The probability of a man hitting a target is $\displaystyle \frac14.$How many times must he fire so that the probability of his hitting the target at least once is greater than$\displaystyle \frac23$ ?

    Answer: -
    3.818 times i-e approximately 4 times
    A man must fire 4 times so that the probability of his hitting the target at least once is greater than $\displaystyle \frac23$
    Verify the answer, if found wrong reply me.
    Answer is not available in the university paper set.So I don't know whether I solved the problem wrongly or correctly.
    Last edited by Vinod; Dec 25th 2011 at 03:41 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Binomial distribution

    Quote Originally Posted by Vinod View Post
    This problem have been taken from the old paper of MASTER of COMMERCE Examination of well-known University in India.
    The probability of a man hitting a target is $\displaystyle \frac14.$How many times must he fire so that the probability of his hitting the target at least once is greater than$\displaystyle \frac23$ ?

    Answer: -
    3.818 times i-e approximately 4 times
    A man must fire 4 times so that the probability of his hitting the target at least once is greater than $\displaystyle \frac23$
    Verify the answer, if found wrong reply me.
    Answer is not available in the university paper set.So I don't know whether I solved the problem wrongly or correctly.
    By definition it must be $\displaystyle (\frac{3}{4})^n<\frac{1}{3} \implies (\frac{4}{3})^n>3 \implies n\ \ln \frac{4}{3}>\ln 3 \implies$

    $\displaystyle \implies n> \frac{\ln 3}{\ln \frac{4}{3}} = 3.8188416793...$



    Marry Christmas from Serbia

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Dec 25th 2011 at 09:20 AM.
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