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Thread: Binomial Probably Variant help

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    Binomial Probably Variant help




    Ok so what i'm thinking here is that the P(X=3) = [(2 choose 2) X (7 choose 2) ] / 10!


    Am I on the right track here?
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    Re: Binomial Probably Variant help

    The denominator should be \binom {10} {5} instead of 10!.

    Note that 10! corresponding to the permutation of 10 distinct objects, while you have two groups of object in which they are indistinguishable within group.

    Therefore you need to reduce the duplicated one by dividing (5!)^2 which result in the answer.

    Actually the formula can be interpret as

    \Pr\{X = x\} = \frac {\displaystyle \binom {x-1} {2} \binom {1} {1} \binom {10-x} {2}} {\displaystyle \binom {10} {5}}, ~~ x = 3, \ldots, 8

    and you may verify that they actually sum to 1.

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    Re: Binomial Probably Variant help

    When x=3 the string has to start MMM.
    When x=8 the string has to end MMM, so by symmetry P(x=3) = P(x=8)
    Likewise P(x=4) = P(x=7) and P(x=5) = P(x=6)
    That should save some calculations.

    I'm pretty sure \binom{10-x}{3} \mbox{ should be} \binom{10-x}{2}

    When x=8 the original formula gives \binom{2}{3}

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    Re: Binomial Probably Variant help


    Yes you are correct. Sorry that is a typo as I post in a rush. Now it is corrected. Thanks.

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