Question 471927
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This is a binomial process.  There are only two possible outcomes on any given trial:  You either pick a guy or a girl.  There are 14 guys out of 20 people (0.7 probability) and there are 6 girls out of 20 people (0.3) probability.  The probabilities are PER TRIAL.


You want to know the probability of exactly two males out of two attempts.


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\)\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)!}]


For your problem:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ p\ =\ 0.7]


Hint:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(n\cr n\right\)\ =\ \left(n\cr 0\right\)\ =\ 1\ \forall\ n\ >\ 0\ \in\ \mathbb{Z}]


And


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^0\ =\ 1\ \forall\ x\ \in\ \mathbb{R}]



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