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Difference between revisions of "2006 AIME A Problems/Problem 5"

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== Solution ==
 
== Solution ==
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For now, assume that face F has a 6 on it and that the face opposite F has a 1 on it.  Let A(n) be the probability of rolling a number n on one die and let B(n) be the probability of rolling a number n on the other die.  One way of getting a 7 is to get a 2 on die A and a 5 on die B.  The probability of this happening is <math>A(2)*B(5)=\frac{1}{6}*\frac{1}{6}=\frac{1}{36}=\frac{8}{288}</math>.  Conversely, one can get a 7 by getting a 2 on die B and a 5 on die A, the probability of which is also <math>\frac{8}{288}</math>.  Getting 7 with a 3 on die A and a 4 on die B also has a probability of <math>\frac{8}{288}</math>, as does getting a 7 with a 4 on die A and a 3 on die B.  Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves a <math>\frac{15}{288}=\frac{5}{96}</math> chance of getting a 1 on die A and a 6 on die B or a 6 on die A and a 1 on die B:
+
For now, assume that face <math>F</math> has a 6 on it so that the face opposite <math>F</math> has a 1 on it.  Let <math>A(n)</math> be the probability of rolling a number <math>n</math> on one die and let <math>B(n)</math> be the probability of rolling a number <math>n</math> on the other die.  One way of getting a 7 is to get a 2 on die <math>A</math> and a 5 on die <math>B</math>.  The [[probability]] of this happening is <math>A(2)\cdot B(5)=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}=\frac{8}{288}</math>.  Conversely, one can get a 7 by getting a 2 on die <math>B</math> and a 5 on die <math>A</math>, the probability of which is also <math>\frac{8}{288}</math>.  Getting 7 with a 3 on die <math>A</math> and a 4 on die <math>B</math> also has a probability of <math>\frac{8}{288}</math>, as does getting a 7 with a 4 on die <math>A</math> and a 3 on die <math>B</math>.  Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves a <math>\frac{15}{288}=\frac{5}{96}</math> chance of getting a 1 on die <math>A</math> and a 6 on die <math>B</math> or a 6 on die <math>A</math> and a 1 on die <math>B</math>:
  
−
<math>A(6)*B(1)+B(6)*A(1)=\frac{5}{96}</math>
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<math>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</math>
  
−
Since both die are the same, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math>:
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Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so
  
−
<math>A(1)*A(6)+A(1)*A(6)=\frac{5}{96}</math>
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<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math>
  
−
<math>2*A(1)*A(6)=\frac{5}{96}</math>
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<math>2\cdot A(1)\cdot A(6)=\frac{5}{96}</math>
  
−
<math>A(1)*A(6)=\frac{5}{192}</math>
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<math>A(1)\cdot A(6)=\frac{5}{192}</math>
  
 
But we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that <math>\sum_{n=1}^6 A(n)=1</math>, so:
 
But we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that <math>\sum_{n=1}^6 A(n)=1</math>, so:
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<math>A(1)=\frac{1}{3}-A(6)</math>
 
<math>A(1)=\frac{1}{3}-A(6)</math>
  
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<math>A(6)*(\frac{1}{3}-A(6))=\frac{5}{192}</math>
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<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math>
  
 
<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
 
<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
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<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
 
<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
  
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<math>A(6)=\frac{-\frac{-1}{3}\pm\sqrt{\frac{-1}{3}^2-4*1*\frac{5}{192}}}{2*1}</math>
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<math>A(6)=\frac{-\frac{-1}{3}\pm\sqrt{\frac{-1}{3}^2-4\cdot1\cdot\frac{5}{192}}}{2\cdot1}</math>
  
 
<math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{9}-\frac{5}{48}}}{2}</math>
 
<math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{9}-\frac{5}{48}}}{2}</math>

Revision as of 15:28, 2 August 2006

Problem

When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than 1/6, the probability of obtaining the face opposite is less than 1/6, the probability of obtaining any one of the other four faces is 1/6, and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is 47/288. Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$


Solution

For now, assume that face $F$ has a 6 on it so that the face opposite $F$ has a 1 on it. Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. One way of getting a 7 is to get a 2 on die $A$ and a 5 on die $B$. The probability of this happening is $A(2)\cdot B(5)=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}=\frac{8}{288}$. Conversely, one can get a 7 by getting a 2 on die $B$ and a 5 on die $A$, the probability of which is also $\frac{8}{288}$. Getting 7 with a 3 on die $A$ and a 4 on die $B$ also has a probability of $\frac{8}{288}$, as does getting a 7 with a 4 on die $A$ and a 3 on die $B$. Subtracting all these probabilities from $\frac{47}{288}$ leaves a $\frac{15}{288}=\frac{5}{96}$ chance of getting a 1 on die $A$ and a 6 on die $B$ or a 6 on die $A$ and a 1 on die $B$:

$A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}$

Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so

$A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}$

$2\cdot A(1)\cdot A(6)=\frac{5}{96}$

$A(1)\cdot A(6)=\frac{5}{192}$

But we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that $\sum_{n=1}^6 A(n)=1$, so:

$A(1)+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+A(6)=\frac{6}{6}$

$A(1)+A(6)+\frac{4}{6}=\frac{6}{6}$

$A(1)+A(6)=\frac{2}{6}$

$A(1)+A(6)=\frac{1}{3}$

Now, combine the two equations:

$A(1)=\frac{1}{3}-A(6)$

$A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}$

$\frac{A(6)}{3}-A(6)^2=\frac{5}{192}$

$A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0$

$A(6)=\frac{-\frac{-1}{3}\pm\sqrt{\frac{-1}{3}^2-4\cdot1\cdot\frac{5}{192}}}{2\cdot1}$

$A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{9}-\frac{5}{48}}}{2}$

$A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{16}{144}-\frac{15}{144}}}{2}$

$A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{144}}}{2}$

$A(6)=\frac{\frac{1}{3}\pm\frac{1}{12}}{2}$

$A(6)=\frac{\frac{4}{12}\pm\frac{1}{12}}{2}$

$A(6)=\frac{\frac{5}{12}}{2}, \frac{\frac{3}{12}}{2}$

$A(6)=\frac{5}{24}, \frac{3}{24}$

$A(6)=\frac{5}{24}, \frac{1}{8}$

We know that $A(6)>\frac{1}{6}$, so it can't be $\frac{1}{8}$. Therefore, it has to be $\frac{5}{24}$ and the answer is $5+24=29$.

See also

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