Question 320800
<font face="Garamond" size="+2">


In order for the result of rolling one die to be divisible by 3, the result of rolling the die must be either a 3 or a 6.  Since there are six possible results and only two of them are considered a success, the probability of success on <i>any one trial</i> is *[tex \Large \frac{2}{6}\ =\ \frac{1}{3}]


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)\ =\ \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)!}]


The probability of <i>at least</i> *[tex \Large k] successes in *[tex \Large n] trials is equal to 1 minus the probability of <i>at most</i> *[tex \Large k\,-\, 1] successes in *[tex \Large n] trials, or:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_n(\geq k)\ =\ 1\ -\ P_n(\leq k\,-\,1)\ =\ \ 1\ -\ \sum_{i\,=\,0}^{k\,-\,1} \left(n\cr i\right\)\left(p\right)^i\left(1\,-\,p\right)^{n\,-\,i}]


So for part a) you need:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(\geq 1)\ =\ 1\ -\ P_{10}(\leq 0)\ =\ 1\ -\ \sum_{i\,=\,0}^{0} \left(10\cr\ 0\right\)\left(\frac{1}{3}\right)^i\left(\frac{2}{3}\right)^{10\,-\,i}\ =\ 1\ -\ \frac{10!}{0!(10)!}\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^{10}] 
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \approx\ 1\ -\ (1)(1)(0.017)\ =\ 0.983]



Or in the common vernacular, highly likely.


For part b) you need:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_{10}(2)\ =\ \left(10\cr \ 2\right\)\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^{8}]


And you can do your own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>