均匀点集间距 I

设抽样点排序为 $0<X_{(1)}<\cdots<X_{(101)}<1$。

条件“任意两点距离都 $>d$”等价于相邻间距都 $>d$（令 $d=1/1000$）：

$$X_{(i+1)}-X_{(i)}>d,\quad i=1,\dots,100.$$

做平移变换 $Y_i=X_{(i)}-(i-1)d$，则

$$0<Y_1<\cdots<Y_{101}<1-100d.$$

该有序区域体积为 $\frac{(1-100d)^{101}}{101!}$；而 101 个点无序抽样的密度对称，乘回 $101!$ 得概率

$$(1-100d)^{101}.$$

代入 $d=1/1000$ 得

$$\left(\frac{9}{10}\right)^{101}.$$

因此 $a=9,b=10,c=101$，$a+b+c=\boxed{120}$。

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英文解析

Let the sampling points be sorted as $0<X_{(1)}<\cdots<X_{(101)}<1$.

The condition "Distance between any two points $>d$" is equivalent to adjacent intervals both $>d$ (order $d=1/1000$):

$$X_{(i+1)}-X_{(i)}>d,\quad i=1,\dots,100.$$

Translate $Y_i=X_{(i)}-(i-1)d$, then

$$0<Y_1<\cdots<Y_{101}<1-100d.$$

The volume of the ordered region is $\frac{(1-100d)^{101}}{101!}$; and the density of the 101-point disordered sample is symmetric, multiplied by the probability of $101!$

$$(1-100d)^{101}.$$

Substitute $d=1/1000$ to get

$$\left(\frac{9}{10}\right)^{101}.$$

So $a=9,b=10,c=101$, $a+b+c=\boxed{120}$.