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

m (Solution 2)
m (Solution 3 (Alcumus))
 
(5 intermediate revisions by 4 users not shown)
Line 21: Line 21:
 
0 &= 192 \left(A(6)\right)^2 - 64 \left(A(6)\right) + 5\\
 
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*}</cmath>
 
A(6)&=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192} =\frac{5}{24}, \frac{1}{8}\end{align*}</cmath>
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{029}</math>.
+
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>.
  
 
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.
Line 28: Line 28:
 
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 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:
 
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*((\frac{1}{6})^2)+2(\frac{1}{6}+x)(\frac{1}{6}-x)=\frac{47}{288}</cmath>
+
<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:
 
multiplying by 288 we get:
<cmath>32+16(6x-1)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15</cmath>
+
<cmath>32+16(1-6x)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15</cmath>
 
dividing by 16 and rearranging we get:
 
dividing by 16 and rearranging we get:
 
<cmath>\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}</cmath>
 
<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{029}</math>
+
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
 +
 
 +
<cmath>\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*}</cmath>
 +
 
 +
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 ==

Latest revision as of 12:10, 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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png