Difference between revisions of "2006 Alabama ARML TST Problems/Problem 10"
(New page: ==Problem== Let <math>p</math> be the probability that Scooby Doo solves any given mystery. The probability that Scooby Doo solves 1800 out of 2006 given mysteries is the same as the proba...) |
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Revision as of 11:26, 17 April 2008
Problem
Let 
be the probability that Scooby Doo solves any given mystery. The probability that
Scooby Doo solves 1800 out of 2006 given mysteries is the same as the probability that he
solves 1801 of them. Find the probability that Scooby Doo solves the mystery of why Eddie
Murphy decided to stop being funny.
Solution
![\[\binom{2006}{1800}\cdot p^{1800}(1-p)^{206}=\binom{2006}{1801}\cdot p^{1801} (1-p)^{205}\]](http://latex.artofproblemsolving.com/3/f/b/3fb81bd6840f081837a77cda200db5995b0bee2d.png)
We want . We solve:
![\[\dfrac{2006!\cdot p^{1800}(1-p)^{206}}{1800!\cdot 206!}=\dfrac{2006!\cdot p^{1801} (1-p)^{205}}{1801!\cdot 205!}\]](http://latex.artofproblemsolving.com/a/8/9/a89994b8fd951ad9950b72ba989edf9e36ede5a7.png)
![\[\dfrac{(1-p)}{206}=\dfrac{p}{1801}\]](http://latex.artofproblemsolving.com/6/5/f/65f6d0337099f31be846977d0760e6d1308ef757.png)
![\[1801-1801p=206p\]](http://latex.artofproblemsolving.com/5/1/8/51899241949e303e0571dfed58064c98fb549c91.png)
![\[1801=2007p\]](http://latex.artofproblemsolving.com/4/e/f/4ef7ff737104541d8ae5de00a719677261921de4.png)
![\[p=\boxed{\dfrac{1801}{2007}}\]](http://latex.artofproblemsolving.com/6/1/f/61fd93194221bcaf6ea2d231ff623c896d77ebe8.png)
