圆上 100 点落在同一半圆内的概率

Points on a Circle II

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

在圆周上均匀随机选取 $n$ 个点。对 $n=100$ ，求这 100 个点都落在同一个半圆内的概率。

答案形如 $\frac{a}{b^c}$ （ $b$ 最小）。求 $a+b+c$ 。

There are $n$ points selected uniformly at randomly around a circle. What is the probability that all $n$ points are on the same semicircle for $n = 100$ ? The answer is in the form $\frac{a}{b^c}$ for integers $a,b,c > 0$ with $b$ minimal. Find $a + b + c$ .

解析

经典结果： $n$ 个点都在同一半圆内的概率为

\frac{n}{2^{n-1}}.

代入 $n=100$ 得 $\frac{100}{2^{99}}$ 。

因此 $a=100,b=2,c=99$ ， $a+b+c=201$ 。

Original Explanation

First consider we already have selected the $n$ points on the circle. Choose one of those points, let's call it $A$ . We draw a line through $A$ and $C$ , the center of the circle; this diameter forms two semicircles. We now only consider the semicircle starting from $A$ in the counterclockwise direction to prevent overcounting.

Then, each of the $n-1$ remaining points has a $\frac{1}{2}$ chance of being within that semicircle. Repeating this for all $n$ points, we find that the solution is simply $\frac{n}{2^{n-1}}$ . Plugging in $n = 100$ , we find our answer to be $\frac{100}{2^{99}}$ . Hence, by plugging in a, b, and c we get the answer to our question is $100 + 2 + 99 = 201$ .