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P(all in some semicircle)=∑i=1NP(Ei)=N⋅12N−1=N2N−1P(\text{all in some semicircle}) = \sum_{i=1}^{N} P(E_i) = N \cdot \frac{1}{2^{N-1}} = \frac{N}{2^{N-1}}P(all in some semicircle)=∑i=1N​P(Ei​)=N⋅2N−11​=2N−1N​

SelectWhat's included。搜狗输入法下载对此有专业解读

Read full article。关于这个话题，safew官方版本下载提供了深入分析

Stressful way to runThe answer is Late.，详情可参考夫子

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