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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 47/288. 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 on it so that the face opposite <math>F</math> has a 1 on it. 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. One way of getting a 7 is to get a 2 on die <math>A</math> and a 5 on die <math>B</math>. The [[probability]] of this happening is <math>A(2)\cdot B(5)=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36}=\frac{8}{288}</math>. Conversely, one can get a 7 by getting a 2 on die <math>B</math> and a 5 on die <math>A</math>, the probability of which is also <math>\frac{8}{288}</math>. Getting 7 with a 3 on die <math>A</math> and a 4 on die <math>B</math> also has a probability of <math>\frac{8}{288}</math>, as does getting a 7 with a 4 on die <math>A</math> and a 3 on die <math>B</math>. Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves a <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>2\cdot 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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| − | But we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that <math>\sum_{n=1}^6 A(n)=1</math>, so:
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| − | <math>A(1)+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+\frac{1}{6}+A(6)=\frac{6}{6}</math>
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| − | <math>A(1)+A(6)+\frac{4}{6}=\frac{6}{6}</math>
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| − | <math>A(1)+A(6)=\frac{2}{6}</math>
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| − | <math>A(1)+A(6)=\frac{1}{3}</math>
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| − | Now, combine the two equations:
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| − | <math>A(1)=\frac{1}{3}-A(6)</math>
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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>A(6)=\frac{-\frac{-1}{3}\pm\sqrt{\frac{-1}{3}^2-4\cdot1\cdot\frac{5}{192}}}{2\cdot1}</math>
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| − | <math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{9}-\frac{5}{48}}}{2}</math>
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| − | <math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{16}{144}-\frac{15}{144}}}{2}</math>
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| − | <math>A(6)=\frac{\frac{1}{3}\pm\sqrt{\frac{1}{144}}}{2}</math>
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| − | <math>A(6)=\frac{\frac{1}{3}\pm\frac{1}{12}}{2}</math>
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| − | <math>A(6)=\frac{\frac{4}{12}\pm\frac{1}{12}}{2}</math>
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| − | <math>A(6)=\frac{\frac{5}{12}}{2}, \frac{\frac{3}{12}}{2}</math>
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| − | <math>A(6)=\frac{5}{24}, \frac{3}{24}</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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| − | *[[2006 AIME II Problems]]
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| − | [[Category:Intermediate Combinatorics Problems]]
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