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

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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>
  
−
<math>\begin{eqnarray*}
+
<math>A(6)=\frac{-\frac{-1}{3}+/-\sqrt{\frac{-1}{3}^2-4*1*\frac{5}{192}}}{2*1}</math>
−
A(6)=\frac{-\frac{-1}{3}+/-\sqrt{\frac{-1}{3}^2-4*1*\frac{5}{192}}}{2*1}
+
 
−
&=&\frac{5}{24}, /frac{1}{8}
+
<math>A(6)=\frac{5}{24}, /frac{1}{8}</math>
−
\end{eqnarray*}</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>.

Revision as of 13:20, 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:

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

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

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

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

$A(6)=\frac{-\frac{-1}{3}+/-\sqrt{\frac{-1}{3}^2-4*1*\frac{5}{192}}}{2*1}$

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

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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