Question 351398
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The probability of *[tex \Large k] successes in *[tex \Large n] trials where the probability of success is *[tex \Large p] 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 computed using *[tex \Large \frac{n!}{k!(n\,-\,k)!]


There are 25 people to choose from, 20 parents and 5 teachers, so the probability of picking a teacher on any given trial is *[tex \Large \frac{5}{25}\ =\ 0.2].


For your first part you want to calculate:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ P_4(4,0.2)\ =\ \left(4\cr{4}\right)\left(0.2\right)^4\left(0.8\right)^{0}]


For the second part:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ P_4(2,0.2)\ =\ \left(4\cr{2}\right)\left(0.2\right)^2\left(0.8\right)^{2}]


For the third part:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ P_4(0,0.2)\ =\ \left(4\cr{0}\right)\left(0.2\right)^0\left(0.8\right)^{4}]


For the fourth part:


*[tex \LARGE\ \ \ \ \ \ \ \ \ \ P_4(1,0.2)\ =\ \left(4\cr{1}\right)\left(0.2\right)^1\left(0.8\right)^{3}]


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