Question 344107
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I'll just give you the formula so that you can answer all three yourself:


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)!}]


(1) *[tex \Large n\ =\ 3], *[tex \Large k\ =\ 3], and *[tex \Large p\ =\ \frac{9}{16}]


(2) *[tex \Large n\ =\ 3], *[tex \Large k\ =\ 0], and *[tex \Large p\ =\ \frac{9}{16}]


(3) *[tex \Large n\ =\ 3], *[tex \Large k\ =\ 2], and *[tex \Large p\ =\ \frac{9}{16}]


Computational hints:


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


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


*[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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