Question

掷一个特定的不公平六面骰子，六个面编号为 $1$ ， $2$ ， $3$ ， $4$ ， $5$ ，和 $6$ ，出现面 $F$ 的概率大于 $\frac{1}{6}$ ，出现与面 $F$ 相对的面的概率小于 $\frac{1}{6}$ ，出现其他每个面的概率都是 $\frac{1}{6}$ ，且每一对相对面的数字之和都是 $7$ 。掷两个这样的骰子时，得到点数和为 $7$ 的概率是 $\frac{47}{288}$ 。已知出现面 $F$ 的概率为 $\frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质正整数，求 $m + n$ 。

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

答案：29

知识点：骰子（概率）平方差

难度评级：2350

解答：

设出现面 $F$ 的概率为 $\frac{1}{6} + x$ ，则与 $F$ 相对的面概率为 $\frac{1}{6} - x$ （六个概率之和必须为 $1$ ）。由于相对面的数字之和为 $7$ ，点数和为 $7$ 恰好发生在两个骰子显示一对相对面时。在六个有序的和为 $7$ 的结果中，四个只用普通面，两个把 $F$ 与它的相对面配对。因此 $\begin{aligned} \frac{47}{288} &= 4\left(\frac{1}{6}\right)^2 \\ &\quad {}+ 2\left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right) \\ &= \frac{1}{6} - 2x^2. \end{aligned}$

因为 $\frac{1}{6} = \frac{48}{288}$ ，得到 $2x^2 = \frac{1}{288}$ ，所以 $x = \frac{1}{24}$ 。面 $F$ 的概率为 $\frac{1}{6} + \frac{1}{24} = \frac{5}{24}$ ，且 $m + n = 5 + 24 = 29$ 。

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 $\begin{aligned} \frac{47}{288} &= 4\left(\frac{1}{6}\right)^2 \\ &\quad {}+ 2\left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right) \\ &= \frac{1}{6} - 2x^2. \end{aligned}$

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 第 5 题

答案：29

难度评级：2350

解答：

其他年份的第 5 题