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专题
Probability / 概率
难度
L1

题目详情

假设从S={10,9,...,9,10}S = \{ -10, -9, ..., 9, 10\}中均匀随机选择两个整数aabb。求 max(0,a)=min(0,b)\max(0, a) = \min(0,b) 的概率

Suppose that two integers aa and bb are uniformly at random selected from S={10,9,...,9,10}S = \{ -10, -9, ..., 9, 10\}. Find the probability that max(0,a)=min(0,b)\max(0, a) = \min(0,b)

解析

注意a<=0a <= 0,为min(0,b)<=0\min(0,b) <= 0。此外,我们还看到b>=0b >= 0,如max(0,a)>=0\max(0, a) >= 0。因此,问题的解就是a<=0a <= 0b>=0b >= 0的概率。 aabb共有212=44121^2 = 441可供选择的方式。其中,aabb 各有 11 个选择,即 a<=0a <= 0b>=0b >= 0,因此我们有 112212=121441\frac{11^2}{21^2} = \frac{121}{441} 的概率。


Original Explanation

Note that a<=0a <= 0, as min(0,b)<=0\min(0,b) <= 0. Furthermore, we also see that b>=0b >= 0, as max(0,a)>=0\max(0, a) >= 0. Therefore, the solution to the problem is the probability that a<=0a <= 0 and b>=0b >= 0. There are 212=44121^2 = 441 total ways in which aa and bb can be selected. Of those, we have 11 choices for each aa and bb, as a<=0a <= 0 and b>=0b >= 0, leaving us with a probability of 112212=121441\frac{11^2}{21^2} = \frac{121}{441}.