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Math Help - Negative Binomial ?

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    Negative Binomial ?

    A box contains 5 electric devices, 2 of them are damaged. We check them randomly each one at a time, until we find out which ones are damaged. Calculate the mean of the number of checks we need to make.

    I thought it's a Negative Binomial distribution, thus, it should be:
    E(X)=r/p=2/0.4.
    but the correct answer should be 3.5.
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    Quote Originally Posted by WeeG View Post
    A box contains 5 electric devices, 2 of them are damaged. We check them randomly each one at a time, until we find out which ones are damaged. Calculate the mean of the number of checks we need to make.

    I thought it's a Negative Binomial distribution, thus, it should be:
    E(X)=r/p=2/0.4.
    but the correct answer should be 3.5.
    Let N be the random variable number of devices removed until two defectives are obtained.

    Calculate Pr(N = 2), Pr(N = 3), .... Pr(N = 6) and then calculate the expected value of N in the usual way.
    Last edited by mr fantastic; May 3rd 2009 at 12:43 AM.
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    MHF Contributor matheagle's Avatar
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    I assume you don't replace the ones you inspect, hence it's a hypergeo
    If you do replace them, then it's a binomial, not neg binomial
    And I'm not sure what your random variable is here.
    Is X=number you must observe in order to find the second defective?
    Then x=2,3,4,5 only.
    P(X=2)=P(defective on first and second)=(2/5)(1/4).
    P(X=3)=P((one defective on first or second pick) and (defective on third))....
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    Quote Originally Posted by matheagle View Post
    I assume you don't replace the ones you inspect, hence it's a hypergeo
    If you do replace them, then it's a binomial, not neg binomial
    And I'm not sure what your random variable is here.
    Is X=number you must observe in order to find the second defective?
    Then x=2,3,4,5 only.
    [snip]
    I think x = 6 will also be needed.

    eg. first selected device could be damaged, next four selected could be undamaged. So to observe the second damaged device a sixth selection is needed. So a maximum of six checks will be needed.

    Edit: I'm not so sure the hypergeometric distribution can be directly used here - because the order is important, the last selection has to be a damaged device eg. *, *, D, *, D versus *, *, D, D, *. In each case there're been five selections that would get counted by the hypergeometric distribution, but only the first is valid in this question.
    Last edited by mr fantastic; May 3rd 2009 at 02:19 AM.
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    MHF Contributor matheagle's Avatar
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    But there are only 5 devices.
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    Quote Originally Posted by matheagle View Post
    But there are only 5 devices.
    Quite right. My mistake. I'd misread as 5 devices and two defectives (= 7 devices).

    But I'm still not convinced the hypergeometric distribution is directly applicable (I await being exposed as an ignoramus for the second time in my life).
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  7. #7
    MHF Contributor matheagle's Avatar
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    I don't think it's hyper either, only my misbehaving children are hyper
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