返回题库

网格矩形概率

2023×2023 grid probability

专题
Probability / 概率
难度
L4

题目详情

在一个 2023×2023 的方格网中，把“中间那一行”（从 1 开始计数时为第 1012 行）整行涂色。

随机选取一个矩形：方法是随机选择两条不同的水平边界线与两条不同的垂直边界线（从所有网格线中选），由此确定一个轴对齐矩形。

问：所选矩形与该中间行有交集的概率是多少？

英文原题

We have a 2023×2023 grid, and the entire “middle row” (row 1012 if indexing from 1) is shaded. We pick a rectangle (by choosing two distinct horizontal boundaries and two distinct vertical boundaries) at random. Probability that the chosen rectangle intersects that middle row?

解析

总矩形数为 (20242)2\binom{2024}{2}^2（水平网格线 2024 条、竖直网格线 2024 条）。

不与中间行相交的矩形只能完全在中间行之上或之下。由于 2023 为奇数，中间行上方有 1011 行、下方也有 1011 行，各自对应的水平网格线数量为 1012。

因此“不相交”的矩形数为

2⋅(10122)⋅(20242).2\cdot \binom{1012}{2}\cdot \binom{2024}{2}.

所以相交概率为

1−2(10122)(20242)(20242)2=1−2(10122)(20242)=1−10112023=10122023.1-\frac{2\binom{1012}{2}\binom{2024}{2}}{\binom{2024}{2}^2} =1-\frac{2\binom{1012}{2}}{\binom{2024}{2}} =1-\frac{1011}{2023} =\frac{1012}{2023}.

英文解析

The total number of rectangles is (20242)2\binom{2024}{2}^2 (2024 horizontal grid lines, 2024 vertical grid lines).

Rectangles that do not intersect the middle row can only be completely above or below the middle row. Since 2023 is odd, there are 1011 rows above the middle row and 1011 rows below, each corresponding to the number of horizontal grid lines is 1012.

So the number of "disjoint" rectangles is

2⋅(10122)⋅(20242).2\cdot \binom{1012}{2}\cdot \binom{2024}{2}.

So the probability of intersection is

1−2(10122)(20242)(20242)2=1−2(10122)(20242)=1−10112023=10122023.1-\frac{2\binom{1012}{2}\binom{2024}{2}}{\binom{2024}{2}^2} =1-\frac{2\binom{1012}{2}}{\binom{2024}{2}} =1-\frac{1011}{2023} =\frac{1012}{2023}.