Question 326869
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Since there are four 5s in the deck of 52 the probability of getting a 5 dealt to you on one draw is *[tex \Large \frac{4}{52}\ =\ \frac{1}{13}].


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 want exactly 1 success out of 4 trials where the probability of success on one trial is *[tex \Large \frac{1}{13}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4\left(1,\frac{1}{13}\right)\ =\ \left(4\cr 1\right\)\left(\frac{1}{13}\right)^1\left(\frac{12}{13}\right)^{3}]


You can do your own arithmetic.  Hint *[tex \LARGE \left(n\cr 1\right\)\ =\ n].


By the way, note that this is the probability of getting exactly one 5, no more, no less.  If you want to know the probability of <i><b>at least</b></i> one 5, that is another thing altogether.  In that case you could calculate the cumulative probability, that is the probability of one 5 plus the probability of two 5s, and so on:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P_4\left(\geq 1,\frac{1}{13}\right)\ =\ \sum_{i\,=\,1}^4\ \left(4\cr i\right\)\left(\frac{1}{13}\right)^i\left(\frac{12}{13}\right)^{4-i}]


Or you could realize that "at least 1" means "any outcome except 0" and then calculate the probability of getting no 5s at all and then subtracting that number from 1.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ P_4\left(0,\frac{1}{13}\right)\ =\ \left(4\cr 0\right\)\left(\frac{1}{13}\right)^0\left(\frac{12}{13}\right)^{4}]


Hint:  *[tex \LARGE \left(n\cr 0\right\)\ =\ 1] and anything raised to the zero power is 1.


Your other problem is done the same way.  You just need to calculate *[tex \LARGE P_3\left(1,\frac{1}{6}\right)] and *[tex \LARGE P_3\left(2,\frac{1}{6}\right)]


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