Solve binamial equation to numeric answer

Hi All,

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

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

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:

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

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

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

Unfortunately, while the first sum can be written as $\displaystyle (.45+ .255)^{25}$, 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.

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

Quote:

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Originally Posted by Vulpes

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

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

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There is a slight modification that makes it somewhat easier,

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

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

That still is a lot of work by hand. Look at this webpage.

Thank you very much for your help!