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Thread: Inequality query

  1. #1
    Junior Member
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    Dec 2010
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    Inequality query

    Hi,
    I'd be grateful if anyone could elucidate why this

    $\displaystyle \sum_{k=1}^2 \binom{2}{k} (1-p)^k p^{2-k} > \sum_{k=2}^4 \binom{4}{k} (1-p)^k p^{4-k} $

    is equivalent to this

    $\displaystyle \sum_{k=0}^1 \binom{4}{k} (1-p)^k p^{4-k}> \binom{2}{0} (1-p)^0 p^{2}$

    I guess it should be obvious but I can't quite see why. Thanks in advance. MD
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  2. #2
    Grand Panjandrum
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    Re: Inequality query

    Quote Originally Posted by Mathsdog View Post
    Hi,
    I'd be grateful if anyone could elucidate why this

    $\displaystyle \sum_{k=1}^2 \binom{2}{k} (1-p)^k p^{2-k} > \sum_{k=2}^4 \binom{4}{k} (1-p)^k p^{4-k} $

    is equivalent to this

    $\displaystyle \sum_{k=0}^1 \binom{4}{k} (1-p)^k p^{4-k}> \binom{2}{0} (1-p)^0 p^{2}$

    I guess it should be obvious but I can't quite see why. Thanks in advance. MD
    Consider the binomial expansions of $\displaystyle ((1-p)+p)^2$ and $\displaystyle ((1-p)+p)^4$

    CB
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