Question 341421
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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\)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)!}]


Here you want to consider the probability over 14 trials where the probability of success (yes, in this context getting hit by a hurricane would be considered a 'success') is 0.07.


The calculation outlined above gives the probability for a specific number of successes.  Where you have situation such as in this problem where you are asked for the probability of "at least" some number, then you have to consider the probability of that number, plus the probability of one more, plus the probability of one more than that...up to the probability of success on every trial.


Symbolically, here is what you would need:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{14}\left(\geq{1},0.07\right)\ =\ \sum_{i=1}^{14}\left(14\cr\ i\right\)\left(0.07\right)^i\left(0.93\right)^{14\,-\,i}]


I'm sure you will agree that is uglier than a mud fence, and you may take it on faith that calculating such a thing has all of the entertainment value of watching paint dry.  But fear not!  There is a simpler way.


The only possibility other than getting hit by at least one is to get hit by none at all!  That means the probability of getting hit by at least one PLUS the probability of getting hit by none at all is equal to 1.  In other words:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{14}\left(\geq{1},0.07\right)\ =\ 1\ -\ P_{14}\left(0,0.07)]


And here is the 'so what':


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{14}\left(\geq{1},0.07\right)\ =\ 1\ -\ P_{14}\left(0,0.07\right)\ =\ 1\ -\ \left(14\cr\ 0\right)\left(0.07\right)^0\left(0.93\right)^{14}]


And since *[tex \LARGE \left(n\cr 0\right\)\ =\ 1\ \forall\ n\ \in\ \mathbb{N}] and *[tex \LARGE x^0\ =\ 1\ \forall\ x\ \in\ \mathbb{R}], the above reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{14}\left(\geq{1},0.07\right)\ =\ 1\ -\ P_{14}\left(0,0.07\right)\ =\ 1\ -\ \left(0.93\right)^{14}]


I'll leave you alone to spend a little quality time with your calculator.


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