2006 Alabama ARML TST Problems/Problem 10

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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
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