The same decomposition works in higher dimensions. For NNN random points on a sphere, the probability they all lie in a hemisphere is also N/2N−1N / 2^{N-1}N/2N−1. The argument is identical: anchor a hemisphere at each point, observe that each event has probability 1/2N−11/2^{N-1}1/2N−1, and verify that at most one anchor can work (the complementary cap is at least a full hemisphere, so no other anchor's hemisphere can straddle it). TODO: If I figure out how to add 3d visualizations to this website, I'll cover the 3D case
Keep reading for HK$10What’s included
，这一点在PDF资料中也有详细论述
Bank details, sex and naked people who seem unaware they are being recorded. Behind Meta’s new smart glasses lies a hidden workforce, uneasy about peering into the most intimate parts of other people’s lives.
“脱アメリカ”本当にできる？悩めるヨーロッパの覚悟とは