Question 1163204
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Probability of a bad part for any random selection: *[tex \Large \frac{50}{450}\ =\ \frac{1}{9}]


Probability of at least two successes out of four trials with a probability of success = *[tex \Large \frac{1}{9}]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\(\geq\,2,4,\frac{1}{9}\)\ =\ \sum_{r=2}^4\,\ {{4}\choose{r}}\,(\frac{1}{9}\)^r\,\cdot\,\(\frac{8}{9}\)^{4-r}]


However, the following is equivalent and less arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\(\geq\,2,4,\frac{1}{9}\)\ =\ 1\ -\ P\(\leq\,1,4,\frac{1}{9}\)\ =\ 1\ -\ \sum_{r=0}^1\,{{4}\choose{r}}\,(\frac{1}{9}\)^r\,\cdot\,\(\frac{8}{9}\)^{4-r}]

								
								
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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