Question

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 $\frac{1}{6},$ the probability of obtaining the face opposite face $F$ is less than $\frac{1}{6},$ the probability of obtaining each of the other faces is $\frac{1}{6},$ and the sum of the numbers on each pair of opposite faces is $7.$ When two such dice are rolled, the probability of obtaining a sum of $7$ is $\frac{47}{288}.$ Given that the probability of obtaining face $F$ is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

答案

Solution:

Let the probability of face $F$ be $\frac{1}{6} + x,$ so the face opposite $F$ has probability $\frac{1}{6} - x$ (the six probabilities must sum to $1$ ). Since opposite faces sum to $7,$ a total of $7$ occurs exactly when the two dice show a pair of opposite faces. Of the six ordered pairs that sum to $7,$ four use only ordinary faces, and two pair $F$ with its opposite. Thus $\frac{47}{288} = 4\left(\frac{1}{6}\right)^2 + 2\left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right) = \frac{1}{6} - 2x^2.$

Since $\frac{1}{6} = \frac{48}{288},$ this gives $2x^2 = \frac{1}{288},$ so $x = \frac{1}{24}.$ The probability of face $F$ is $\frac{1}{6} + \frac{1}{24} = \frac{5}{24},$ and $m + n = 5 + 24 = 29.$

2006 AIME II Problem 5

Solution:

Problem 5 in Other Years