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

(New page: == Problem == A coin that comes up heads with probability and tails with probability independently on each flip is flipped eight times. Suppose the probability of three heads and five t...)
 
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== Problem ==
 
== Problem ==
  
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A coin that comes up heads with probability and tails with probability independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to of the probability of five heads and three tails. Let , where and are relatively prime positive integers. Find .
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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>.
  
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== Solution ==
+
== Solution 1 ==
  
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If we let the odds of a tails (1-p) equal t, then the probablity of three heads and five tails is:
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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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p^3t^5
 
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The probability of five heads and three tails is:
 
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p^5t^3
 
  
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25p^3t^5 = p^5t^3
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<cmath>\begin{align*}
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25t^2 = p^2
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25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\
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5t = p
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25(1-p)^2&=p^2 \\
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5(1-p) = p
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25p^2-50p+25&=p^2 \\
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5 - 5p = p
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24p^2-50p+25&=0 \\
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5 = 6p
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p&=\frac {5}{6}\end{align*}</cmath>
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p = 5/6
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 +
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 \\
 +
5-5p&=p \\
 +
5&=6p \\
 +
p&=\frac {5}{6}\end{align*}</cmath>
 +
 
 +
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>
 +
 
 +
Therfore, 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 ==
 +
{{AIME box|year=2009|n=I|num-b=2|num-a=4}}
 +
{{MAA Notice}}

Revision as of 17:13, 3 June 2021

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 $\frac56$

Therfore, 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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