2006 AIME II Problems/Problem 5

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: \[32+16(6x-1)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15\] 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 $\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{29}$

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

Invalid username
Login to AoPS