Difference between revisions of "2006 AIME A Problems/Problem 5"
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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=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: | ||
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A(6)*B(1)+B(6)*A(1)=5/96 | A(6)*B(1)+B(6)*A(1)=5/96 | ||
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Since both die are the same, this reduces to: | Since both die are the same, this reduces to: | ||
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2*A(6)*A(1)=5/96 | 2*A(6)*A(1)=5/96 | ||
A(6)*A(1)=5/192 | A(6)*A(1)=5/192 | ||
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But we know that A(2)=A(3)=A(4)=A(5)=1/6, so: | But we know that A(2)=A(3)=A(4)=A(5)=1/6, so: | ||
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A(6)+A(1)=1/3 | A(6)+A(1)=1/3 | ||
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Now, combine the two equations: | Now, combine the two equations: | ||
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A(1)=1/3-A(6) | A(1)=1/3-A(6) | ||
A(6)*(1/3-A(6))=5/192 | A(6)*(1/3-A(6))=5/192 | ||
| − | + | A(6)/3-A(6)^2=5/192 | |
| + | A(6)^2-A(6)/3+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 == | == See also == | ||
Revision as of 17:45, 1 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 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:
A(6)*B(1)+B(6)*A(1)=5/96
Since both die are the same, this reduces to:
2*A(6)*A(1)=5/96 A(6)*A(1)=5/192
But we know that A(2)=A(3)=A(4)=A(5)=1/6, so:
A(6)+A(1)=1/3
Now, combine the two equations:
A(1)=1/3-A(6) A(6)*(1/3-A(6))=5/192 A(6)/3-A(6)^2=5/192 A(6)^2-A(6)/3+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.
