# Solve binamial equation to numeric answer

• Jun 8th 2013, 09:12 AM
Vulpes
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
• Jun 8th 2013, 10:00 AM
HallsofIvy
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.
• Jun 8th 2013, 12:07 PM
Vulpes
Re: Solve binamial equation to numeric answer
Hi HallsofIvy,

So the only way of calculating the second sum is by actually summing the terms by hand?

Kind regards,
Chris
• Jun 8th 2013, 12:43 PM
Plato
Re: Solve binamial equation to numeric answer
Quote:

Originally Posted by Vulpes
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.
• Jun 8th 2013, 01:42 PM
Vulpes
Re: Solve binamial equation to numeric answer
Thank you very much for your help!