Difference between revisions of "2006 AIME II Problems/Problem 5"
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== Problem == | == Problem == | ||
− | + | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face <math> F </math> is greater than <math>1/6</math>, the probability of obtaining the face opposite is less than <math>1/6</math>, the probability of obtaining any one of the other four faces is <math>1/6</math>, 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 <math>47/288</math>. Given that the [[probability]] of obtaining face <math> F </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] positive integers, find <math> m+n. </math> | |
− | When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face <math> F </math> 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 <math>47/288</math>. Given that the [[probability]] of obtaining face <math> F </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] positive integers, find <math> m+n. </math> | ||
== Solution == | == Solution == | ||
+ | [[Without loss of generality]], assume that face <math>F</math> has a 6, so the opposite face has a 1. Let <math>A(n)</math> be the probability of rolling a number <math>n</math> on one die and let <math>B(n)</math> be the probability of rolling a number <math>n</math> on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of <math>\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}</math>, totaling <math>4 \cdot \frac{1}{36} = \frac{1}{9}</math>. Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves <math>\frac{15}{288}=\frac{5}{96}</math> chance of getting a 1 on die <math>A</math> and a 6 on die <math>B</math> or a 6 on die <math>A</math> and a 1 on die <math>B</math>: | ||
− | + | <cmath>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</cmath> | |
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Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so | Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so | ||
− | + | <cmath>\begin{align*}A(1)\cdot A(6)+A(1)\cdot A(6)&=\frac{5}{96}\\ | |
− | + | A(1)\cdot A(6)&=\frac{5}{192}\end{align*}</cmath> | |
Also, we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that the total probability must be <math>1</math>, so: | Also, we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that the total probability must be <math>1</math>, so: | ||
− | + | <cmath>A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6} \Longrightarrow A(1)+A(6)=\frac{1}{3}</cmath> | |
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Combining the equations: | Combining the equations: | ||
− | + | <cmath>\begin{align*}A(6)\left(\frac{1}{3}-A(6)\right)&=\frac{5}{192}\\ | |
− | + | 0 &= 192 \left(A(6)\right)^2 - 64 \left(A(6)\right) + 5\\ | |
− | + | A(6)&=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192} =\frac{5}{24}, \frac{1}{8}\end{align*}</cmath> | |
− | + | We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>. Therefore, the probability is <math>\frac{5}{24}</math> and the answer is <math>5+24=\boxed{029}</math>. | |
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− | We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>. | ||
Note also that the initial assumption that face <math>F</math> was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was. | Note also that the initial assumption that face <math>F</math> was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was. | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | [[Category:Intermediate Probability Problems]] |
Revision as of 13:12, 26 April 2008
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 , the probability of obtaining the face opposite is less than , the probability of obtaining any one of the other four faces is , 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 . Given that the probability of obtaining face is where and are relatively prime positive integers, find
Solution
Without loss of generality, assume that face has a 6, so the opposite face has a 1. Let be the probability of rolling a number on one die and let be the probability of rolling a number on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of , totaling . Subtracting all these probabilities from leaves chance of getting a 1 on die and a 6 on die or a 6 on die and a 1 on die :
Since the two dice are identical, and so
Also, we know that and that the total probability must be , so:
Combining the equations:
We know that , so it can't be . Therefore, the probability is and the answer is .
Note also that the initial assumption that face was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |