求救:一道用面积求概率的数学题(英文的)

来源:百度知道 编辑:UC知道 时间:2024/06/02 01:49:00
Two circles of radius 1 are to be constructed as follows. The center of circle A is chosen uniformly and at random from the line segment joining(0,0) and(2,0) . The center of circle B is chosen uniformly and at random, and independently of the first choice, from the line segment joining(0,1) to(2,1) . What is the probability that circles A and B intersect?

解答:Circles centered at A and B will overlap if A and B are closer to each other than if the circles were tangent. The circles are tangent when the distance between their centers is equal to the sum of their radii. Thus, the distance from A to B will be2 . Since and A areB separated by 1 vertically, they must be separated by 根号3 horizontally. Thus, if A与B距离小于根号3 the circles intersect.

Now, plot the two random variables AX and BX on the coordinate plane. Each variable ranges from 0 to2 . The circles intersect if the variables are within 根3 of each other. Thus, the area in which the circles don't intersect

题目想必你看懂了 ,就不再翻译
当两圆外切的时候,AB=2
A,B两点的水平距离=√3
所以两圆相交的概率即A,B两点的水平距离小于√3的概率
令A,B的横坐标分别为x,y,都在[0,2]上任意变化,所以在坐标平面上表现为一个边长为2的正方形,总面积看做概率1
|x-y|<√3(-√3<x-y<√3)表示的区域与正方形重叠的面积即符合要求的部分
它比上总面积为概率

如果是按你中文说的,把个边变长的概率看作1,面积的概率也为1,对边长进行积分就可以了 ~~