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

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(→‎Solution)
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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><math>A(1)+A(2)+A(3)+A(4)+A(5)+A(6)=1</math></math>
 
  
 
<math>A(1)+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+A(6)=\frac{6}{6}</math>
 
<math>A(1)+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+A(6)=\frac{6}{6}</math>
  
 
<math>A(1)+A(6)+\frac{4}{6}=\frac{6}{6}</math>
 
<math>A(1)+A(6)+\frac{4}{6}=\frac{6}{6}</math>
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<math>A(1)+A(6)=\frac(2}{6}</math>
 
  
 
<math>A(6)+A(1)=\frac{1}{3}</math>
 
<math>A(6)+A(1)=\frac{1}{3}</math>
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Now, combine the two equations:
 
Now, combine the two equations:
  
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A(1)=<math>\frac{1}{3}</math>-A(6)
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<math>A(1)=\frac{1}{3}-A(6)</math>
  
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A(6)*(<math>\frac{1}{3}</math>-A(6))=<math>\frac{5}{192}</math>
+
<math>A(6)*(\frac{1}{3}-A(6))=\frac{5}{192}</math>
  
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<math>\frac{A(6)}{3}</math>-A(6)^2=<math>\frac{5}{192}</math>
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<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
  
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A(6)^2-<math>\frac{A(6)}{3}</math>+<math>\frac{5}{192}</math>=0
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<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
  
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A(6)=5/24, 1/8
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<math>A(6)=5/24, 1/8</math>
  
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We know that A(6)>1/6, so it can't be 1/8.  Therefore, it has to be 5/24 and the answer is 5+24=29.
+
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>.
  
 
== See also ==
 
== See also ==

Revision as of 13:12, 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 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 $A(2)*B(5)=\frac{1}{6}*\frac{1}{6}=\frac{1}{36}=\frac{8}{288}$. 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:

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

Since both die are the same, $B(1)=A(1)$ and $B(6)=A(6)$:

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

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

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

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:

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

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

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

Now, combine the two equations:

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

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

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

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

$A(6)=5/24, 1/8$

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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