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

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== Problem ==
 
== Problem ==
  
−
A coin that comes up heads with probability <math>p > 0</math> and tails with probability <math>1 - p > 0</math> independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to <math>\frac {1}{25}</math> of the probability of five heads and three tails. Let <math>p = \frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
+
A coin that comes up heads with probability <math>p > 0</math> and tails with probability <math>1 - p > 0</math> independently on each flip is flipped <math>8</math> times. Suppose that the probability of three heads and five tails is equal to <math>\frac {1}{25}</math> of the probability of five heads and three tails. Let <math>p = \frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
−
== Solution ==
+
== Solution 1 ==
  
 
The probability of three heads and five tails is <math>\binom {8}{3}p^3(1-p)^5</math> and the probability of five heads and three tails is <math>\binom {8}{3}p^5(1-p)^3</math>.
 
The probability of three heads and five tails is <math>\binom {8}{3}p^3(1-p)^5</math> and the probability of five heads and three tails is <math>\binom {8}{3}p^5(1-p)^3</math>.
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25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\
 
25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\
 
25(1-p)^2&=p^2 \\
 
25(1-p)^2&=p^2 \\
 +
25p^2-50p+25&=p^2 \\
 +
24p^2-50p+25&=0 \\
 +
p&=\frac {5}{6}\end{align*}</cmath>
 +
 +
Therefore, the answer is <math>5+6=\boxed{011}</math>.
 +
 +
== Solution 2 ==
 +
 +
We start as shown above. However, when 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*}
 
5(1-p)&=p \\
 
5(1-p)&=p \\
 
5-5p&=p \\
 
5-5p&=p \\
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p&=\frac {5}{6}\end{align*}</cmath>
 
p&=\frac {5}{6}\end{align*}</cmath>
  
−
Therefore, the answer is <math>5+6=\boxed{011}</math>.
+
Now, we get <math>5+6=\boxed{011}</math>.
 +
 
 +
~Jerry_Guo
 +
 
 +
== Solution 3 ==
 +
Rewrite it as : <math>(P)^3</math><math>(1-P)^5=\frac {1}{25}</math> <math>(P)^5</math><math>(1-P)^3</math>
 +
 
 +
This can be simplified as <math>24P^2 -50P + 25 = 0</math>
 +
 
 +
This can be factored into <math>(4P-5)(6P-5)</math>
 +
 
 +
This yields two solutions: <math>\frac54</math> (ignored because it would result in <math>1-p<0</math>  ) or <math>\frac56</math>
 +
 
 +
Therefore, the answer is <math>5+6</math> = <math>\boxed {011}</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/NL79UexadzE
 +
 
 +
~IceMatrix
 +
 
 +
==Video Solution 2==
 +
https://www.youtube.com/watch?v=P00iOJdQiL4
 +
 
 +
~Shreyas S
  
 
== 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}}

Latest revision as of 02:35, 16 January 2022

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 1

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, when 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 $24P^2 -50P + 25 = 0$

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

This yields two solutions: $\frac54$ (ignored because it would result in $1-p<0$ ) or $\frac56$

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

Video Solution

https://youtu.be/NL79UexadzE

~IceMatrix

Video Solution 2

https://www.youtube.com/watch?v=P00iOJdQiL4

~Shreyas S

See also

2009 AIME I (Problems • Answer Key • Resources)
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Problem 4
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All AIME Problems and Solutions

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