Question 984389
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The probability of *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any individual trial is *[tex \Large p] is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k,p)\ =\ {{n}\choose{k}}\left(p\right)^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \LARGE {{n}\choose{k}}] is the number of combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is calculated by *[tex \Large \frac{n!}{k!(n\,-\,k)!}]


Then the probability of less than or equal to *[tex \Large k] successes (in other words *[tex \Large k] successes or fewer), in *[tex \Large n] trials with probability *[tex \Large p] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(\leq{k},p)\ =\ \sum_{i=0}^k\,{{n}\choose{i}}\left(p\right)^i\left(1\,-\,p\right)^{n\,-\,i}]


You have some arithmetic to do.  On the other hand, if you have Excel on your machine (or Numbers if you are using a Mac), open a spreadsheet and type:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =BINOMDIST(3,5,0.25,TRUE)]


And the spreadsheet will dutifully give you the answer to more decimal places than you could ever need as soon as you hit "Enter".


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \