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
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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==
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<cmath>p=\boxed{\dfrac{1801}{2007}}</cmath>
 
<cmath>p=\boxed{\dfrac{1801}{2007}}</cmath>
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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==
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{{ARML box|year=2006|state=Alabama|num-b=9|num-a=11}}

Latest revision as of 20:29, 20 May 2012

Problem

Let $p$ 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

\[\binom{2006}{1800}\cdot p^{1800}(1-p)^{206}=\binom{2006}{1801}\cdot p^{1801} (1-p)^{205}\]

We want $p$. We solve:

\[\dfrac{2006!\cdot p^{1800}(1-p)^{206}}{1800!\cdot 206!}=\dfrac{2006!\cdot p^{1801} (1-p)^{205}}{1801!\cdot 205!}\]

\[\dfrac{(1-p)}{206}=\dfrac{p}{1801}\]

\[1801-1801p=206p\]

\[1801=2007p\]

\[p=\boxed{\dfrac{1801}{2007}}\]

Note: Both $p = 0$ and $p = 1$ 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
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