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Math Help - Probability, where am I going wrong?

  1. #1
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    Probability, where am I going wrong?

    An urn contains 10 white and 5 red marbles. Three marbles are selected at random without replacement from the urn. Set up the probability distribution for the randomn variable X defined as the number of red marbles selected. Determine the expected value and the variance of X.

    So what I did

    P(WWW) = 0.2637
    P(WWR) = 0.1648
    P(WRW) = 0.1648
    P(WRR) = 0.0733
    P(RWW) = 0.1648
    P(RWR) = 0.0733
    P(RRW) = 0.0733
    P(RRR) = 0.0220

    The mean is 1 and the variance is 0.5056.

    My problem is, I keep on getting 0.5714, where am I going wrong?
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  2. #2
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    I think that you are going about in the hard way.
    P(R=x)=\frac{\binom{10-x}{3-x}\binom{5}{x}}{\binom{15}{3}}, That is the probability of having x=~0,~1,~2,\text{ or }3 red balls in the sample.

    BTW: I get the same variance as you.
    Last edited by Plato; May 19th 2010 at 01:56 PM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    This is a hypergeometric distribution.
    You can check the mean (1) and variance at...http://en.wikipedia.org/wiki/Hyperge...c_distribution

    The mean is the same as with the binomial (np), even though this is sampling wor.
    n=3 and p=5/15=1/3 giving you E(X)=1.
    The variances are slightly different, but as the lot sizes increase they approach each other.
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