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随机折两刀能组成三角形的概率

Probability of Triangle

专题
Probability / 概率
难度
L4

题目详情

长度为 1 的木棍在 [0,1][0,1] 上随机选取两个点切开（独立均匀）。问三段能组成三角形的概率是多少？

A stick of length 1 is cut at two random points (chosen independently and uniformly on [0,1][0,1]). What is the probability that the three resulting segments form a triangle?

解析

答案是 14\frac{1}{4}。

设切点为 0<x<y<10<x<y<1，三段长度为 x, y−x, 1−yx,\ y-x,\ 1-y。

组成三角形等价于最长段小于 1/21/2，对应区域面积占比为 1/41/4，因此概率为 1/41/4。

Original Explanation

Let the cut points be x,yx,y with x<yx<y. The segment lengths are x, y−x, 1−y.x,\,y-x,\,1-y. They form a triangle if:

  • x+(y−x)>1−y  ⇒  y>12,x + (y - x) > 1 - y \;\Rightarrow\; y>\tfrac12,
  • x+(1−y)>y−x  ⇒  y<12+x,x + (1 - y) > y - x \;\Rightarrow\; y<\tfrac12+x,
  • (y−x)+(1−y)>x  ⇒  x<12.(y - x) + (1 - y) > x \;\Rightarrow\; x<\tfrac12.

The region satisfying all conditions occupies 14\tfrac{1}{4} of the half-square (0<x<y<1)(0<x<y<1), so the probability is 14\tfrac14.