Question 349766
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Can't answer this specifically because you didn't share the number of trials.


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\)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 \frac{n!}{k!(n\,-\,k)!}]


You have *[tex \Large p\ =\ 0.15] and you have *[tex \Large k] is at least 1.  But you didn't give *[tex \Large n].


The probability of at least 1 is the probability of exactly 1 plus the probability of exactly 2 plus exactly 3...and so on up to the probability of exactly *[tex \Large n] out of *[tex \Large n].  Such a calculation would look like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(\geq{1},0.15)\ =\ \sum_{i=1}^n\left(n\cr i\right\)\left(0.15\right)^i\left(0.85\right)^{n\,-\,i}]


The calculation of which, depending on the size of *[tex \Large n], could range anywhere from mildly uncomfortable to excruciating.  Ah, but take heart, Grasshopper!  There is a better way.


Note that if "at least 1" is one possibility, then the only other possiblity is "none at all".  Hence the probability of "at least one" plus the probability of "none at all" must equal 1.  Therefore, if you calculate the probability of none at all and subtract that from 1, you get your desired probability.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(\geq{1},0.15)\ =\ 1\ -\ P_n(0,0.15)\ =\ 1\ -\ \left(n\cr 0\right\)\left(0.15\right)^0\left(0.85\right)^n].


But since *[tex \LARGE \left(n\cr 0\right\)\ =\ 1] and *[tex \LARGE x^0\ =\ 1], this all reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(\geq{1},0.15)\ =\ 1\ -\ P_n(0,0.15)\ =\ 1\ -\ \left(0.85\right)^n].


All you need to do is determine *[tex \Large n] and then do the arithmetic.


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