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

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(→‎Solution)
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== Solution ==
 
== 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)=1/6*1/6=1/36=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 8/288.  Getting 7 with a 3 on die A and a 4 on die B also has a probability of 8/288, as does getting a 7 with a 4 on die A and a 3 on die B.  Subtracting all these probabilities from 47/288 leaves a 15/288=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:
+
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)=1/6*1/6=<math>\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:
  
−
A(6)*B(1)+B(6)*A(1)=5/96
+
A(6)*B(1)+B(6)*A(1)=<math>\frac{5}{96}</math>
  
−
Since both die are the same, this reduces to:
+
Since both die are the same, B(1)=A(1) and B(6)=A(6):
  
−
2*A(6)*A(1)=5/96
+
A(6)*A(1)+A(6)*A(1)=<math>\frac{5}{96}</math>
  
−
A(6)*A(1)=5/192
+
2*A(6)*A(1)=<math>\frac{5}{96}</math>
  
−
But we know that A(2)=A(3)=A(4)=A(5)=1/6, so:
+
A(6)*A(1)=<math>\frac{5}{192}</math>
  
−
A(6)+A(1)=1/3
+
But we know that A(2)=A(3)=A(4)=A(5)=<math>\frac{1}{6}</math>, so:
 +
 
 +
A(6)+A(1)=<math>\frac{1}{3}</math>
  
 
Now, combine the two equations:
 
Now, combine the two equations:
  
−
A(1)=1/3-A(6)
+
A(1)=<math>\frac{1}{3}</math>-A(6)
  
−
A(6)*(1/3-A(6))=5/192
+
A(6)*(<math>\frac{1}{3}</math>-A(6))=<math>\frac{5}{192}</math>
  
−
A(6)/3-A(6)^2=5/192
+
<math>\frac{A(6)}{3}</math>-A(6)^2=<math>\frac{5}{192}</math>
  
−
A(6)^2-A(6)/3+5/192=0
+
A(6)^2-<math>\frac{A(6)}{3}</math>+<math>\frac{5}{192}</math>=0
  
 
A(6)=5/24, 1/8
 
A(6)=5/24, 1/8

Revision as of 13:43, 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)=1/6*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)=$\frac{1}{6}$, so:

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)>1/6, so it can't be 1/8. Therefore, it has to be 5/24 and the answer is 5+24=29.

See also

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