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

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== Problem ==
 
== Problem ==
 +
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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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>
+
== 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>:
  
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== Solution ==
+
<cmath>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</cmath>
−
 
 
−
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>:
 
−
 
 
−
<math>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</math>
 
  
 
Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so
 
Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so
  
−
:<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math>
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<cmath>\begin{align*}A(1)\cdot A(6)+A(1)\cdot A(6)&=\frac{5}{96}\\
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:<math>A(1)\cdot A(6)=\frac{5}{192}</math>
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A(1)\cdot A(6)&=\frac{5}{192}\end{align*}</cmath>
  
 
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:
 
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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<cmath>A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6} \Longrightarrow A(1)+A(6)=\frac{1}{3}</cmath>
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:<math>A(1)+A(6)=\frac{1}{3}</math>
 
  
 
Combining the equations:
 
Combining the equations:
  
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:<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math>
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<cmath>\begin{align*}A(6)\left(\frac{1}{3}-A(6)\right)&=\frac{5}{192}\\
−
 
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0 &= 192 \left(A(6)\right)^2 - 64 \left(A(6)\right) + 5\\
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:<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math>
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A(6)&=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192} =\frac{5}{24}, \frac{1}{8}\end{align*}</cmath>
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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, the probability is <math>\frac{5}{24}</math> and the answer is <math>5+24=\boxed{29}</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>
+
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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:<math>A(6)=\frac{5}{24}, \frac{1}{8}</math>
+
== 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 \cdot \left(\frac{1}{6} \right)^2+2 \left(\frac{1}{6}+x \right) \left(\frac{1}{6}-x \right)=\frac{47}{288}</cmath>
 +
multiplying by 288 we get:
 +
<cmath>32+16(1-6x)(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{29}</math>
 +
== Solution 3 (Alcumus) ==
 +
Let <math>p(a,b)</math> denote the probability of obtaining <math>a</math> on the first die and <math>b</math> on the second. Then the probability of obtaining a sum of 7 is<cmath>p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).</cmath>Let the probability of obtaining face <math>F</math> be <math>(1/6)+x</math>. Then the probability of obtaining the face opposite face <math>F</math> is <math>(1/6)-x</math>. Therefore
  
−
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>.
+
<cmath>\begin{align*}
 +
{{47}\over{288}}&=
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4\left({1\over6}\right)^2+2\left({1\over6}+x\right)
 +
\left({1\over6}-x\right)\cr&=
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{4\over36}+2\left({1\over36}-x^2\right)\cr&=
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{1\over6}-2x^2.
 +
\end{align*}</cmath>
  
−
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.
+
Then <math>2x^2=1/288</math>, and so <math>x=1/24</math>. The probability of obtaining face <math>F</math> is therefore <math>(1/6)+(1/24)=5/24</math>, and <math>m+n=\boxed{29}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
[[Category:Intermediate Combinatorics Problems]]
 +
[[Category:Intermediate Probability Problems]]
 +
{{MAA Notice}}

Latest revision as of 11:10, 5 November 2023

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 $5+24=\boxed{29}$.

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. 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: \[4 \cdot \left(\frac{1}{6} \right)^2+2 \left(\frac{1}{6}+x \right) \left(\frac{1}{6}-x \right)=\frac{47}{288}\] multiplying by 288 we get: \[32+16(1-6x)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15\] 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 $\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{29}$

Solution 3 (Alcumus)

Let $p(a,b)$ denote the probability of obtaining $a$ on the first die and $b$ on the second. Then the probability of obtaining a sum of 7 is\[p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).\]Let the probability of obtaining face $F$ be $(1/6)+x$. Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$. Therefore

\begin{align*} {{47}\over{288}}&= 4\left({1\over6}\right)^2+2\left({1\over6}+x\right) \left({1\over6}-x\right)\cr&= {4\over36}+2\left({1\over36}-x^2\right)\cr&= {1\over6}-2x^2. \end{align*}

Then $2x^2=1/288$, and so $x=1/24$. The probability of obtaining face $F$ is therefore $(1/6)+(1/24)=5/24$, and $m+n=\boxed{29}$.

See also

2006 AIME II (Problems • Answer Key • Resources)
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