Question 990166
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The probability of *[tex \Large k] successes in *[tex \Large n] trials where *[tex \Large p] is the probability of success on any given trial 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)!}]


For the first one, you need 3 of 9 plus 4 of 9 plus 5 of 9 plus 6 of 9 where p = 0.25.


More than 5 is 6, 7, 8, or 9.


Get out your calculator or if you have Excel:
<pre>
=BINOMDIST(6,9,0.25,TRUE) - BINOMDIST(2,9,0.25,TRUE)

and

=1-BINOMDIST(5,9,0.25,TRUE)
</pre>

By the way, these same formulas work in Numbers on a Mac


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

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