AMC 8 2023 Problem 23
AMC8 2023 年第 23 题
概率★★★★☆
Each square in a grid is randomly filled with one of the 4 gray and white tiles shown below on the right. What is the probability that the tiling will contain a large gray diamond in one of the smaller grids? Below is an example of such a tiling.
将3×3方格表中的每个小方格随机的用如下图左侧所示的4个灰白双色方块之一嵌入，有一个大灰色菱形将出现在某个2×2子方格表中的概率是多少？一个这样的镶嵌方案的例子如下图右侧所示，
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
C
解析 · Solution
解法1
常规解法 · 数有利情形
共 种等概率。某 2×2 成菱形需四格灰三角都朝中心 ⇒ 这 4 格唯一确定、余 5 格随意 。3×3 有 4 个 2×2 子格、互不重叠，有利数 。概率 。
巧解 · 单格概率 ×4
指定一个 2×2 成菱形的概率 （四格各 朝中心）。4 个位置且互斥 ⇒ 。
巧解 · 期望线性
记 4 个 2×2 子格为 ，令 表示 恰拼成大菱形、否则 。单个子格成菱形需四格各自朝中心，。所求 ，由期望的线性（不要求事件互斥）
比解法2 更稳：无需论证四个位置两两不相交。
Solution 1
Elementary · count favorable cases
All tilings are equally likely. For a given 2×2 block to form the large diamond, its four tiles must each point to the center — that fixes those 4 tiles, with the other 5 free (). The 3×3 grid has 4 non-overlapping 2×2 blocks, so the favorable count is , and the probability is .
Slick · single-square probability ×4
A specified 2×2 block forms the diamond with probability (each of its 4 tiles points to the center). With 4 mutually exclusive positions, the answer is .
Slick · linearity of expectation
Let be the four 2×2 blocks and if forms the diamond, else 0. Each needs its four tiles pointing to the center, so . The required probability is , and by linearity of expectation (no disjointness needed)
This is safer than the previous method: no need to argue the four positions are pairwise disjoint.
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