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

Revision as of 19:20, 4 July 2013

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 $\binom {8}{3}p^5(1-p)^3$ .

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

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

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 == See also ==
 {{AIME box|year=2009|n=I|num-b=2|num-a=4}}
+{{MAA Notice}}

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

Revision as of 19:20, 4 July 2013

Problem

Solution

See also