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

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== Problem ==
 
== Problem ==
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When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face <math> F </math> 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 <math> F </math>  is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers, find <math> m+n. </math>  
+
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face <math> F </math> 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 <math>47/288</math>. Given that the [[probability]] of obtaining face <math> F </math>  is <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] positive integers, find <math> m+n. </math>  
  
  
 
== Solution ==
 
== Solution ==
−
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>:
+
For now, assume that face <math>F</math> has a 6, so the opposite face has a 1.  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. 7 can be obtained by rolling a <math>A(n)=2</math> and <math>B(n)=5</math>, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of <math>\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}</math>, totaling <math>4 \cdot \frac{1}{36} = \frac{1}{9}</math>. Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves <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)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</math>
 
<math>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</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
 
Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so
  
−
<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math>
+
:<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math>
 +
:<math>A(1)\cdot A(6)=\frac{5}{192}</math>
  
−
<math>2\cdot A(1)\cdot A(6)=\frac{5}{96}</math>
+
Also, we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that the total probability must be <math>1</math>, so:
  
−
<math>A(1)\cdot A(6)=\frac{5}{192}</math>
+
:<math>A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}</math>
 +
:<math>A(1)+A(6)=\frac{1}{3}</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:
+
Combining the equations:
  
−
<math>A(1)+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+A(6)=\frac{6}{6}</math>
+
:<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math>
  
−
<math>A(1)+A(6)+\frac{4}{6}=\frac{6}{6}</math>
+
:<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
  
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<math>A(1)+A(6)=\frac{2}{6}</math>
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:<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
  
−
<math>A(1)+A(6)=\frac{1}{3}</math>
+
:<math>\displaystyle 192 A(6)^2 - 64 A(6) + 5 = 0</math>
  
−
Now, combine the two equations:
+
:<math>A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}</math>
  
−
<math>A(1)=\frac{1}{3}-A(6)</math>
+
:<math>A(6)=\frac{64\pm16}{384}</math>
  
−
<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math>
+
:<math>A(6)=\frac{5}{24}, \frac{1}{8}</math>
−
 
 
−
<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
 
−
 
 
−
<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
 
−
 
 
−
<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{16}{144}-\frac{15}{144}}}{2}</math>
 
−
 
 
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<math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{144}}}{2}</math>
 
−
 
 
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<math>A(6)=\frac{\frac{1}{3}\pm\frac{1}{12}}{2}</math>
 
−
 
 
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<math>A(6)=\frac{\frac{4}{12}\pm\frac{1}{12}}{2}</math>
 
−
 
 
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<math>A(6)=\frac{\frac{5}{12}}{2}, \frac{\frac{3}{12}}{2}</math>
 
−
 
 
−
<math>A(6)=\frac{5}{24}, \frac{3}{24}</math>
 
−
 
 
−
<math>A(6)=\frac{5}{24}, \frac{1}{8}</math>
 
  
 
We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>.  Therefore, it has to be <math>\frac{5}{24}</math> and the answer is <math>5+24=29</math>.
 
We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>.  Therefore, it has to be <math>\frac{5}{24}</math> and the answer is <math>5+24=29</math>.
−
 
  
 
Note also that the initial assumption that face <math>F</math> was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was.
 
Note also that the initial assumption that face <math>F</math> was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was.
  
 
== See also ==
 
== See also ==
−
*[[2006 AIME II Problems]]
+
{{AIME box|year=2006|n=II|num-b=4|num-a=6}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 17:56, 21 February 2007

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, so the opposite face has a 1. 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. 7 can be obtained by rolling a $A(n)=2$ and $B(n)=5$, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$, totaling $4 \cdot \frac{1}{36} = \frac{1}{9}$. Subtracting all these probabilities from $\frac{47}{288}$ leaves $\frac{15}{288}=\frac{5}{96}$ 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

$A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}$
$A(1)\cdot A(6)=\frac{5}{192}$

Also, we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that the total probability must be $1$, so:

$A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}$
$A(1)+A(6)=\frac{1}{3}$

Combining the equations:

$A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}$
$\frac{A(6)}{3}-A(6)^2=\frac{5}{192}$
$A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0$
$\displaystyle 192 A(6)^2 - 64 A(6) + 5 = 0$
$A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}$
$A(6)=\frac{64\pm16}{384}$
$A(6)=\frac{5}{24}, \frac{1}{8}$

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 $5+24=29$.

Note also that the initial assumption that face $F$ was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was.

See also

2006 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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