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Difference between revisions of "2006 AIME II Problems/Problem 5"

 
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#REDIRECT [[2006 AIME A Problems/Problem 5]]
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== Problem ==
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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 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>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 Algebra Problems]]

Revision as of 18:02, 25 September 2007

Problem

When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ 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 $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

For now, assume that face $F$ has a 6, so the opposite face has a 1. 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. 7 can be obtained by rolling a $A(n)=2$ and $B(n)=5$, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$, totaling $4 \cdot \frac{1}{36} = \frac{1}{9}$. Subtracting all these probabilities from $\frac{47}{288}$ leaves $\frac{15}{288}=\frac{5}{96}$ 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 $B$:

$A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}$

Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so

$A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}$
$A(1)\cdot A(6)=\frac{5}{192}$

Also, we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that the total probability must be $1$, so:

$A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}$
$A(1)+A(6)=\frac{1}{3}$

Combining the equations:

$A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}$
$\frac{A(6)}{3}-A(6)^2=\frac{5}{192}$
$A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0$
$192 A(6)^2 - 64 A(6) + 5 = 0$
$A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}$
$A(6)=\frac{64\pm16}{384}$
$A(6)=\frac{5}{24}, \frac{1}{8}$

We know that $A(6)>\frac{1}{6}$, so it can't be $\frac{1}{8}$. Therefore, it has to be $\frac{5}{24}$ and the answer is $5+24=29$.

Note also that the initial assumption that face $F$ 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
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