Question

掷两个公平骰子，每个骰子至少有 $6$ 个面。每个骰子的各面分别标着从 $1$ 到该骰子面数的不同整数。掷出和为 $7$ 的概率是掷出和为 $10$ 的概率的 $\dfrac34$ ，且掷出和为 $12$ 的概率是 $\dfrac{1}{12}$ 。两个骰子的面数总和最小可能是多少？

Two fair dice, each with at least $6$ faces are rolled. On each face of each die is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum of $7$ is $\dfrac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\dfrac{1}{12}.$ What is the least possible number of faces on the two dice combined?

$16$

$17$

$18$

$19$

$20$

答案：B

知识点：骰子（概率）最优化

难度评级：2120

解答：

设两个骰子的面数为 $a\le b$ 。因为两个骰子都至少有 $6$ 个面，和为 $7$ 恰有 $6$ 种方式，所以和为 $10$ 有 $6\div\tfrac34=8$ 种方式。

掷出 $10$ 的方式数为 $\min(a,9)$ $-\max(1,10-b)+1=8$ 。和为 $12$ 的概率为 $\tfrac{1}{12}$ ，所以它有 $\tfrac{ab}{12}$ 种方式。

试 $8$ ：和为 $10$ 有 $a\ge8$ 种方式，和为 $12$ 有 $b\ge9$ 种方式。两个条件都满足，得到 $a+b\ge17$ 。检查所有更小的总和 $(a,b)=(8,9)$ 均不成立，所以最小值为 $17$ 。 $10$ $8$ $6$ $3,4,\ldots,8$ 。 $6/(8\cdot9)=1/12$ ，

所以正确答案是 B。

Let the dice have $a\le b$ faces. Since both have at least $6$ faces, a sum of $7$ occurs in exactly $6$ ways, so a sum of $10$ occurs in $6\div\tfrac34=8$ ways.

The number of ways to roll $10$ is $\min(a,9)$ $-\max(1,10-b)+1=8.$ A sum of $12$ has probability $\tfrac{1}{12},$ so it occurs in $\tfrac{ab}{12}$ ways.

Having $8$ outcomes for a sum of $10$ requires $a\ge8$ and $b\ge9,$ so necessarily $a+b\ge17.$ For $(a,b)=(8,9),$ a sum of $10$ has $8$ outcomes, while a sum of $12$ has the $6$ outcomes with the first die showing $3,4,\ldots,8.$ Since $6/(8\cdot9)=1/12,$ both conditions hold and the lower bound $17$ is attained.

Thus, the correct answer is B.

2021 AMC 12B Spring 第 19 题

答案：B

难度评级：2120

解答：

其他年份的第 19 题