Question 438855
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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)\ =\ \left(n\cr k\right\)\left(p\right)^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 \frac{n!}{k!(n\,-\,k)!}]


But you need the probability of getting 24 OR MORE correct.  So you need the probability of exactly 24 plus the probability of exactly 25...exactly 26...and so on.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_38\left(\geq24,0.5\right)\ =\ \sum_{i=24}^{38}\ \left(n\cr i\right\)\left(0.5\right)^i\left(0.5\right)^{38\,-\,i}]


This is a long and arduous calculation, so get out your calculator and start cranking.


On the other hand, if you have Microsoft Excel or Numbers for Mac you can open a spreadsheet and put in the following formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =1-\text{BINOMDIST}(23,38,0.5,\text{TRUE})]


Round appropriately and you will have your answer.  The BINOMDIST part of that will give you the probability of getting zero plus P(1) plus P(2), and so on up to 23.  The TRUE part of that will give you the cumulative sum from 0 through 23. If you put in FALSE, you would get the probability of 23 exactly.  1 minus the result of the cumulative amount will give you the probability of 24 or more.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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