Question 1160252
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The probability of at least one is the probability of exactly one, P(1), plus the probability of exactly two, P(2), plus P(3), plus P(4), ..., plus P(20).


Each of the listed probabilities is calculated with the formula:


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


Where *[tex \Large k] is the desired number of successes, *[tex \Large n] is the total number of trials, and *[tex \Large p] is the probability of success on any given trial.


So the entire sum is:


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


and for your problem *[tex \Large k\,=\,1], *[tex \Large n\,=\,20], and *[tex \Large p\,=\,0.4]


As should be readily apparent, this calculation would require a tremendous amount of tedious arithmetic. However, note that the probability of at least one success plus the probability of zero successes is equal to 1.  Hence, the probability of at least one is equal to 1 minus the probability of zero successes, to wit:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(\geq{1},20,0.4)\ =\ 1\ -\ P(0,20,0.4)\ = 1\ - \ {{20}\choose{0}}\,(0.4)^0\(0.6\)^{20}]


Which reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ (1)(1)(0.6)^{20}]


And I leave you with that little bit of calculator work.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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