Question 1199818
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The probability of exactly *[tex \Large k] successes in *[tex \Large n] independent trials with a uniform probability of success on any given trial of *[tex \Large p] is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(n,k,p)\ =\ {{n}\choose{k}}\,(p)^k(1\,-\,p)^{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.


The probability of at most *[tex \Large k] successes in *[tex \Large n] independent trials with a uniform probability of success on any given trial of *[tex \Large p] is given by the sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(n,\leq\,k,p)\ =\ \sum_{r=0}^k\,{{n}\choose{r}}\,(p)^r(1\,-\,p)^{n-r}]


Which is the probability of exactly zero successes plus the probability of exactly one and so on up to the probability of exactly *[tex \Large k] successes.


The probability of at least *[tex \Large k] successes in *[tex \Large n] independent trials with a uniform probability of success on any given trial of *[tex \Large p] is given by the sum:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(n,\geq\,k,p)\ =\ \sum_{r=k}^n\,{{n}\choose{r}}\,(p)^r(1\,-\,p)^{n-r}]


Which is the probability of exactly *[tex \Large k] successes plus the probability of exactly *[tex \Large k\,+\,1] and so on up to the probability of exactly n successes.


Just plug in your numbers, remembering to express the probability as a decimal, and do the indicated arithmetic.


It may save you some arithmetic if you recognize that *[tex \Large P(n,\leq\,k,p)\ =\ 1\ -\ P(n,\geq\,k+1,p)]

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

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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