Question 1113461
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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}}\,\cdot\,\(p\)^k\,\cdot\,\(1\,-\,p\)^{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)!}].


To answer the given question you need to calculate the probability of getting exactly 9 out of 10 questions correct and the probability of getting all 10 out of 10 questions correct when the probability of getting any single question correct is 0.50 and then find the sum of those two probabilities.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(k>8,10,0.5)\ =\ {{10}\choose{9}}\,\cdot\,\(.5\)^9\,\cdot\,\(1\,-\,0.5\)^{10\,-\,9}\ +\ {{10}\choose{10}}\,\cdot\,\(.5\)^{10}\,\cdot\,\(1\,-\,0.5\)^{10\,-\,10}]


A couple of facts that will make the arithmetic simpler:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  {{n}\choose{n-1}}\ =\ n]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  {{n}\choose{n}}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a^0\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  a^1\ =\ 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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