Question 331633
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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)!}]


Your problem is that you want the probability of "at most 4" tails in 5 flips.  That means you need to calculate the probability of 0 tails in 5 flips and add that to the probability of 1 tail in 5 flips, 2 tails in 5...you get the picture.  Each flip has a probability of 0.5 for a tail, so your calculation will look like this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5(\leq4,0.5)\ =\ \sum_{i=0}^4\left(5\cr i\right\)(0.5)^i(0.5)^{5\,-\,i}]


Which calculation is guaranteed to provide all the entertainment value that can be obtained watching paint dry.


But all is not lost.  Notice that the only alternative to "at most 4 tails" in 5 flips is all 5 of the flips coming up tails.  That means we can simply find the probability of 5 tails out of 5 flips and subtract from 1 to get our answer.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5(\leq4,0.5)\ =\ 1\ -\ P_5(5,0.5)\ =\ 1\ -\ \left(5\cr 5\right\)(0.5)^5(0.5)^{0}]


And this looks MUCH better, particularly when you realize that *[tex \LARGE \left(n\cr n\right\)\ =\ 1\ \forall\ n\ \in\ \mathbb{N}] and *[tex \LARGE x^0\ =\ 1\ \forall\ x\ \in\ \mathbb{R}], so really this reduces to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_5(\leq4,0.5)\ =\ 1\ -\ P_5(5,0.5)\ =\ 1\ -\ (0.5)^5\ =\ 1\ -\ \frac{1}{32}\ \approx\ 97%]


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