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Math Help - Solve binamial equation to numeric answer

  1. #1
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    Question Solve binamial equation to numeric answer

    Hi All,

    In my textbook there is the following binomial expression which is solved to a numerical answer:

    \sum_{i=13}^{25} {25 \choose i} (0.45)^{i} (0.55)^{25-i} \approx 0.306

    I'd like to know if there is a way to calculate this result without having to calculate the entire summation.
    I know that from the binomial theorem:

    \sum_{k=0}^{n} {n \choose k} x^{n-k} y^{k} = (x+y)^{n}

    But how can I use this to calculate the value above? Because the lower bound of the sum in the textbook example is not 0, I cannot us the binomial theorem straight away. Is there some way in which I can rewrite the formula to make use of the binomial theory?

    Kind regards,
    Chris
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  2. #2
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    Re: Solve binamial equation to numeric answer

    If the lower bound is not 0, "add and subtract" to get that:
    \sum_{i= 13}^{25}\begin{pmatrix}25 \\ i\end{pmatrix}(0.45)^I(0.255)^{25- i}= \sum_{i= 0}^{25}\begin{pmatrix}25 \\ i\end{pmatrix}(0.45)^i(0.255)^{25- i}-  \sum_{i= 0}^{12}\begin{pmatrix}25 \\ i\end{pmatrix}(0.45)^i(0.255)^{25- i}

    Unfortunately, while the first sum can be written as (.45+ .255)^{25}, the second sum cannot be written in that way because the upper term in the binomial coefficient is NOT the same as the highest value of k.
    Last edited by HallsofIvy; June 8th 2013 at 10:09 AM.
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  3. #3
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    Re: Solve binamial equation to numeric answer

    Hi HallsofIvy,

    Thank you for your quick answer.
    So the only way of calculating the second sum is by actually summing the terms by hand?

    Kind regards,
    Chris
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  4. #4
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    Re: Solve binamial equation to numeric answer

    Quote Originally Posted by Vulpes View Post
    In my textbook there is the following binomial expression which is solved to a numerical answer:

    \sum_{i=13}^{25} {25 \choose i} (0.45)^{i} (0.55)^{25-i} \approx 0.306

    There is a slight modification that makes it somewhat easier,

    Because \sum_{i=0}^{25} {25 \choose i} (0.45)^{i} (0.55)^{25-i}=1

    Then \sum_{i=13}^{25} {25 \choose i} (0.45)^{i} (0.55)^{25-i}=1-\sum_{i=0}^{12} {25 \choose i} (0.45)^{i} (0.55)^{25-i}.

    That still is a lot of work by hand. Look at this webpage.
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  5. #5
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    Re: Solve binamial equation to numeric answer

    Thank you very much for your help!
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