Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2006 AIME A Problems/Problem 5

Revision as of 13:00, 2 August 2006 by Ninja glace (talk | contribs) (→‎Solution)

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:

$A(1)+A(2)+A(3)+A(4)+A(5)+A(6)=1$

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

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

$A(1)+A(6)=\frac(2}{6}$ (Error compiling LaTeX. Unknown error_msg)

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_A_Problems/Problem_5&oldid=9046"