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Difference between revisions of "2009 AIME I Problems/Problem 3"

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Therefore, the answer is <math>5+6=\boxed{011}</math>.
 
Therefore, the answer is <math>5+6=\boxed{011}</math>.
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==Video Solution==
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https://youtu.be/NL79UexadzE
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~IceMatrix
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2009|n=I|num-b=2|num-a=4}}
 
{{AIME box|year=2009|n=I|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:50, 19 February 2020

Problem

A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

The probability of three heads and five tails is $\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\binom {8}{3}p^5(1-p)^3$.

\begin{align*} 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ 25(1-p)^2&=p^2 \\ 25p^2-50p+25=p^2 \\ 24p^2-50p+25=0 \\ p&=\frac {5}{6}\end{align*}

Therefore, the answer is $5+6=\boxed{011}$.

Video Solution

https://youtu.be/NL79UexadzE

~IceMatrix

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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