Question 1037745
<font face="Times New Roman" size="+2">


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)!}]



You want the probability of 8 successes out of 10 trials where the probability of success on any given trial is 0.15, so:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(8,0.15)\ =\ {{10}\choose{8}}\left(0.15\right)^8\left(0.85\right)^{2}]


That is just an arithmetic problem.


The probability of at least 1 is the sum of all numbers of successes except 0, so you can use the above process to calculate the probability of exactly 1, then exactly 2, exactly 3, and so on all the way up to 10 and then add all 10 results, OR you can just calculate the probability of exactly 0 successes and subtract that probability from 1.


The choice should be pretty obvious unless you are a glutton for punishment.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(\geq{1},0.15)\ =\ 1\ -\ P_{10}(0,0.15)\ =\ 1\ -\ {{10}\choose{0}}\left(0.15\right)^0\left(0.85\right)^{10}]


Note that *[tex \Large {{n}\choose{0}}\ =\ 1] for any positive integer *[tex \Large n] and *[tex \Large x^0\ =\ 1] for all real numbers *[tex \Large x] which will simplify your calculation to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(\geq{1},0.15)\ =\ 1\ -\ P_{10}(0,0.15)\ =\ 1\ -\ \left(0.85\right)^{10}]


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 \ \ \ \ \ \ \ \ \ \  

</font>