Question 1164736
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Two things:


1. If the probability of success for any given trial is *[tex \Large p], then the probability of failure for any given trial is *[tex \Large 1\ -\ p].


2. The probability of exactly *[tex \Large k] successes in *[tex \Large n] trials where the probability of success on any given trial is *[tex \Large p] is given by *[tex \Large P(k,n,p)\ =\ {{n}\choose{k}}\,(p)^k(1-p)^{n-k}]


For part a) you need the probability of 9 successes in 14 trials times the probability of success once, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(P(9,14,0.65)\)(p)\ =\ \({{14}\choose{9}}\,(0.65)^9(0.35)^{5}\)(p)]


For part b) it is just *[tex \Large P(10,15,0.65)]


And for part c) you need the probability of missing on the first two tries times the probability of hitting once, i.e. *[tex \Large \(P(2,2,0.35)\)(p)]


You can do your own arithmetic

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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