Question 1166269
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Since engine failures are independent events, the multiplication rule applies and the probability of three failures is therefore *[tex \Large q^3].  And the probability of two failures is *[tex \Large q^2].  Since *[tex \Large q], by definition is less than or equal to 1, and nobody would actually ride in an airplane unless *[tex \Large q] was nearly zero, *[tex \Large q^3] cannot be larger than *[tex \Large q^2], and most likely is considerably smaller.


																
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">


I > Ø
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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