2009 AIME I Problems/Problem 3

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

If we let the odds of a tails $(1-p)$ equal $t$ , then the probability of three heads and five tails is $28{p^3}{t^5}$ The probability of five heads and three tails is $28{p^5}{t^3}$

$25*28{p^3}{t^5} = 28*{p^5}{t^3}$ $25{t^2} = {p^2}$ $5t = p$ $5(1 - p) = p$ $5 - 5p = p$ $5 = 6p$ $p = \frac {5} {6}$ $5 + 6 = \boxed{11}$

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2009 AIME I Problems/Problem 3

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