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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 5]] |
| − | 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>
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| − | == Solution ==
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| − | For now, 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 <math>A(n)=2</math> and <math>B(n)=5</math>, 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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| − | <math>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</math>
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| − | Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so
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| − | :<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math>
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| − | :<math>A(1)\cdot A(6)=\frac{5}{192}</math>
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| − | 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:
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| − | :<math>A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}</math>
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| − | :<math>A(1)+A(6)=\frac{1}{3}</math>
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| − | Combining the equations:
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| − | :<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math>
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| − | :<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
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| − | :<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math>
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| − | :<math>\displaystyle 192 A(6)^2 - 64 A(6) + 5 = 0</math>
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| − | :<math>A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}</math>
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| − | :<math>A(6)=\frac{64\pm16}{384}</math>
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| − | :<math>A(6)=\frac{5}{24}, \frac{1}{8}</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>. Therefore, it has to be <math>\frac{5}{24}</math> and the answer is <math>5+24=29</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.
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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|num-b=4|num-a=6}}
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| − | [[Category:Intermediate Combinatorics Problems]]
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