Question

从一个由单位正方形组成的 $n \times n$ 方格中，不放回地随机选出两个单位正方形。求最小正整数 $n$ ，使得这两个选出的正方形水平相邻或竖直相邻的概率小于 $\frac{1}{2015}$ 。

Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected squares are horizontally or vertically adjacent is less than $\frac{1}{2015}.$

答案

答案：90

知识点：基本概率数对计数不等式

难度评级：2270

解答：

$n$ 行中的每一行有 $n - 1$ 对水平相邻的正方形，共 $n(n-1)$ 对；竖直相邻同样有 $n(n-1)$ 对。在所有 $\binom{n^2}{2} = \frac{n^2(n^2-1)}{2}$ 个等可能的正方形对中，相邻的概率为 $\frac{2n(n-1) \cdot 2}{n^2(n^2 - 1)} = \frac{4}{n(n+1)}.$

需要 $n(n+1) \gt 4 \cdot 2015 = 8060$ 。因为 $89 \cdot 90 = 8010$ ，而 $90 \cdot 91 = 8190$ ，满足条件的最小 $n$ 是 $90$ 。

Each of the $n$ rows contains $n - 1$ horizontally adjacent pairs, so there are $n(n-1)$ horizontal pairs and likewise $n(n-1)$ vertical pairs. Out of $\binom{n^2}{2} = \frac{n^2(n^2-1)}{2}$ equally likely pairs, the probability of adjacency is $\frac{2n(n-1) \cdot 2}{n^2(n^2 - 1)} = \frac{4}{n(n+1)}.$

We need $n(n+1) \gt 4 \cdot 2015 = 8060.$ Since $89 \cdot 90 = 8010$ and $90 \cdot 91 = 8190,$ the least such $n$ is $90.$

2015 AIME II 第 5 题

答案：90

难度评级：2270

解答：

其他年份的第 5 题