Question 448034
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There are two ways to figure this.  Since there are only four possible outcomes for tossing two coins, you can simply enumerate them and count the successes.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ HH\ \ ] Failure: 2 heads
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ HT\ \ ] Success: 1 head
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ TH\ \ ] Success: 1 head
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ TT\ \ ] Failure: 0 heads


Two successes out of four possiblities.  Probablity is 1/2.


Or you can get fancy and use the formula for 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:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(k,p)\ =\ \left(n\cr k\right\)\left(p\right)^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)!}]

Or for your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_2(1,0.5)\ =\ \left(2\cr 1\right\)\left(0.5\right)^1\left(0.5\right)^1\ =\ 2\,\cdot\,0.5\,\cdot\,0.5\ =\ 0.5]





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