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>A(6)=\frac{-\frac{-1}{3} | + | <math>A(6)=\frac{-\frac{-1}{3}\pm\sqrt{\frac{-1}{3}^2-4*1*\frac{5}{192}}}{2*1}</math> |
− | <math>A(6)=\frac{\frac{1}{3} | + | <math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{9}-\frac{5}{48}}}{2}</math> |
− | <math>A(6)=\frac{\frac{1}{3} | + | <math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{16}{144}-\frac{15}{144}}}{2}</math> |
− | <math>A(6)=\frac{\frac{1}{3} | + | <math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{144}}}{2}</math> |
− | <math>A(6)=\frac{\frac{1}{3} | + | <math>A(6)=\frac{\frac{1}{3}\pm\frac{1}{12}}{2}</math> |
− | <math>A(6)=\frac{\frac{4}{12} | + | <math>A(6)=\frac{\frac{4}{12}\pm\frac{1}{12}}{2}</math> |
<math>A(6)=\frac{\frac{5}{12}}{2}, \frac{\frac{3}{12}}{2}</math> | <math>A(6)=\frac{\frac{5}{12}}{2}, \frac{\frac{3}{12}}{2}</math> |
Revision as of 13:32, 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 .