Question 281485
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The probability of *[tex \Large k] successes out of *[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)\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k}]


Where *[tex \Large \left(n\cr k\right)] is the number of ways to select *[tex \Large k] things out of *[tex \Large n] things and is equal to *[tex \Large \frac{n!}{k!(n\,-\,k)!}].


So for the first part of your question, you need to calculate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(8)\ =\ \left(10\cr \,8\right\)\left(\frac{1}{2}\right)^8\left(\frac{1}{2}\right)^{2}]


Because you want the probability of 8 heads in 10 flips where the probability of getting heads is *[tex \Large \frac{1}{2}]


The probability of <b><i>more than</i></b> *[tex \Large k] successes out of *[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)\ =\ \sum_{i=k\,+\,1}^n\left(\left(n\cr i\right\)p^i\left(1\,-\,p\right)^{n\,-\,i}\right)]


So, you need to calculate:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(> 8)\ =\ P_{10}(9)\ +\ P_{10}(10)\ =\  \left(10\cr \,9\right\)\left(\frac{1}{2}\right)^9\left(\frac{1}{2}\right)^{1}\ +\  \left(10\cr \,10\right\)\left(\frac{1}{2}\right)^{10}\left(\frac{1}{2}\right)^{0}]


Use MS Excel to check your work.  Pick any cell on a blank worksheet and type in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{=BINOMDIST(8,10,0.5,FALSE)}]


for the first part and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{=1-BINOMDIST(8,10,0.5,TRUE)}]


for the second part.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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