Question 1113639
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The probability of *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any given trial is *[tex \Large p] is given by:


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


Where *[tex \LARGE {{n}\choose{k}}] is the number of ways to choose *[tex \Large k] things from *[tex \Large n] things calculated by *[tex \LARGE \frac{n!}{k!(n-k)!}]


You want 10 successes in 10 trials where the probability of success on any trial is 0.5 (presuming a fair coin).


Just plug in the numbers *[tex \Large k\ =\ 10,\ \ n\ =\ 10,\ \ p\ =\ 0.5] and do the arithmetic.


There are a couple of facts that will simplify your computations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  {{n}\choose{n}}\ =\ 1] for all positive integers *[tex \Large n]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a^0\ =\ 1] for all real *[tex \Large a]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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