Question 316540
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The basic formula for the probability of exactly *[tex \Large k] successes out of *[tex \Large n] trials where the probability of a successful outcome on a single trial is fixed at *[tex \Large p] 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 combinations of *[tex \Large n] things taken *[tex \Large k] at a time and is calculated by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(n\cr k\right\)\ =\ \frac{n!}{k!(n\ -\ k)!}]


But you need the probability of 15 right answers plus 16 right answers plus ... plus 19 right answers plus 20 right answers.


So you need:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{20}(\geq15)\ =\ \sum_{i=15}^{20}\left(20\cr \ i\right\)0.5^i\left(1\,-\,0.5\right)^{n\,-\,i}\ =\ \sum_{i=15}^{20}\left(20\cr \ i\right\)0.5^i\left(0.5\right)^{n\,-\,i}\  ]


But since *[tex \LARGE 0.5^i\left(0.5\right)^{n\,-\,i}\ =\ 0.5^n], we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{20}(\geq15)\ =\ \left(\left(20\cr 15\right\)+\ \left(20\cr 16\right\)\ +\left(20\cr 17\right\)\ +\left(20\cr 18\right\)\ +\left(20\cr 19\right\)\ +\left(20\cr 20\right\)\right)0.5^{20}]


Get busy.  You have some serious calculator punching to do.  Or you can use Excel.  The formula:  =COMBIN(N,K) where N and K are the values you want gives you the value of *[tex \LARGE \left(n\cr k\right\)]


I ran this out and got a little over 2% probability.


SUPER-DOOPER DOUBLE PLUS EXTRA CREDIT.


10 question multiple-choice exam.  All questions have four possible responses.  Passing grade is 6 or better correct.  Probability you can pass by just guessing?


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