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

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== Solution 2 ==  
 
== Solution 2 ==  
  
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We start as shown above. However, as we get to <math>25(1-p)^2=p^2</math>, we square root both sides to get <math>5(1-p)=p</math>. We can do this because we know that both <math>p</math> and <math>1-p</math> are between 0 and one, so they are positive. Now, we have:  
+
We start as shown above. However, as we get to <math>25(1-p)^2=p^2</math>, we square root both sides to get <math>5(1-p)=p</math>. We can do this because we know that both <math>p</math> and <math>1-p</math> are between <math>0</math> and <math>1</math>, so they are both positive. Now, we have:  
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}

Revision as of 17:12, 2 June 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 $5+6=\boxed{011}$.

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

Video Solution

https://youtu.be/NL79UexadzE

~IceMatrix

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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