Difference between revisions of "2006 AIME A Problems/Problem 5"
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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>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> |
Revision as of 13:29, 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 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 is where and are relatively prime positive integers, find
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 . 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 . Getting 7 with a 3 on die A and a 4 on die B also has a probability of , as does getting a 7 with a 4 on die A and a 3 on die B. Subtracting all these probabilities from leaves a 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:
Since both die are the same, and :
But we know that and that , so:
Now, combine the two equations:
We know that , so it can't be . Therefore, it has to be and the answer is .