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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>
  
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== Solution ==
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== 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.
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== Solution 2 ==
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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>
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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:
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<cmath>4*((\frac{1}{6})^2)+2(\frac{1}{6}+x)(\frac{1}{6}-x)=\frac{47}{288}</cmath>
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multiplying by 288 we get:
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<cmath>32+16(6x-1)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15</cmath>
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dividing by 16 and rearranging we get:
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<cmath>\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}</cmath>
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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 20:35, 17 February 2014

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 1

Without loss of generality, 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 2 and 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

\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*}

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

\[A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6} \Longrightarrow A(1)+A(6)=\frac{1}{3}\]

Combining the equations:

\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*} We know that $A(6)>\frac{1}{6}$, so it can't be $\frac{1}{8}$. Therefore, the probability is $\frac{5}{24}$ and the answer is $5+24=\boxed{029}$.

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. We could have labelled $A(6)$ as $p$, 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 $\frac{1}{6}$, $\frac{1}{6}$, $\frac{1}{6}$, $\frac{1}{6}$ ,$\frac{1}{6}+x$ ,$\frac{1}{6}-x$ 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: \[4*((\frac{1}{6})^2)+2(\frac{1}{6}+x)(\frac{1}{6}-x)=\frac{47}{288}\] multiplying by 288 we get: \[32+16(6x-1)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15\] dividing by 16 and rearranging we get: \[\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}\] so the probability F which is greater than $\frac{1}{6}$ is equal $\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{029}$

See also

2006 AIME II (Problems • Answer Key • Resources)
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