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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==Problem== | ==Problem== | ||
| − | Let <math>p</math> be the probability that Scooby Doo solves any given mystery. The probability that | + | 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 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. |
| − | 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== | ==Solution== | ||
| Line 19: | Line 16: | ||
<cmath>p=\boxed{\dfrac{1801}{2007}}</cmath> | <cmath>p=\boxed{\dfrac{1801}{2007}}</cmath> | ||
| + | |||
| + | Note: Both <math>p = 0</math> and <math>p = 1</math> satisfy the given condition, but based on the context of the problem, these aren't the answers we're looking for. | ||
==See also== | ==See also== | ||
| + | {{ARML box|year=2006|state=Alabama|num-b=9|num-a=11}} | ||
Latest revision as of 19:29, 20 May 2012
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)
Note: Both 
and 
satisfy the given condition, but based on the context of the problem, these aren't the answers we're looking for.
See also
| 2006 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 9 |
Followed by: Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
