# Difference between revisions of "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 $8$ times. Suppose that 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 $5+6=\boxed{011}$.

## Solution 2

We start as shown above. However, as we get to $25(1-p)^2=p^2$, we square root both sides to get $5(1-p)=p$. We can do this because we know that both $p$ and $1-p$ are between $0$ and $1$, so they are both positive. Now, we have:

\begin{align*} 5(1-p)&=p \\ 5-5p&=p \\ 5&=6p \\ p&=\frac {5}{6}\end{align*}

Now, we get $5+6=\boxed{011}$.

~Jerry_Guo

## Solution 3

Rewrite it as : $(P)^3$$(1-P)^5$=$\frac {1}{25}$ $(P)^5$$(1-P)^3$

This can be simplified as $24P^2 -50P + 25 = 0$

The can be factored into: $(4P-5)(6P-5)$

This yields two solutions: $5/4$(ignored) and $\boxed {5/6}$

~IceMatrix

~Shreyas S