Difference between revisions of "2006 AIME II Problems/Problem 5"
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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 <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> | ||
− | == Solution == | + | == Solution 1== |
[[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>: | [[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>: | ||
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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. We could have labelled <math>A(6)</math> as <math>p</math>, for example, and replaced the others with variables too, but the notation would have been harder to follow. | 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. We could have labelled <math>A(6)</math> as <math>p</math>, for example, and replaced the others with variables too, but the notation would have been harder to follow. | ||
+ | |||
+ | == Solution 2 == | ||
+ | we have that the cube probabilities to land on its faces are <math>\frac{1}{6}</math>, <math>\frac{1}{6}</math>, <math>\frac{1}{6}</math>, <math>\frac{1}{6}</math> ,<math>\frac{1}{6}+x</math> ,<math>\frac{1}{6}-x </math> | ||
+ | we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: | ||
+ | <cmath>4*((\frac{1}{6})^2)+2(\frac{1}{6}+x)(\frac{1}{6}-x)=\frac{47}{288}</cmath> | ||
+ | multiplying by 288 we get: | ||
+ | <cmath>32+16(6x-1)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15</cmath> | ||
+ | dividing by 16 and rearranging we get: | ||
+ | <cmath>\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}</cmath> | ||
+ | so the probability F which is greater than <math>\frac{1}{6}</math> is equal <math>\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{029}</math> | ||
== See also == | == See also == |
Revision as of 21:35, 17 February 2014
Contents
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 1
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. We could have labelled as , for example, and replaced the others with variables too, but the notation would have been harder to follow.
Solution 2
we have that the cube probabilities to land on its faces are , , , , , we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: multiplying by 288 we get: dividing by 16 and rearranging we get: so the probability F which is greater than is equal
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 |
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