HMMT 十一月 2023 · 冲刺赛 · 第 6 题

HMMT November 2023 — Guts Round — Problem 6

[6] There are five people in a room. They each simultaneously pick two of the other people in the room independently and uniformly at random and point at them. Compute the probability that there exists a group of three people such that each of them is pointing at the other two in the group. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2023, November 11, 2023 — GUTS ROUND Organization Team Team ID#

解析

[6] There are five people in a room. They each simultaneously pick two of the other people in the room independently and uniformly at random and point at them. Compute the probability that there exists a group of three people such that each of them is pointing at the other two in the group. Proposed by: Neil Shah 5 Answer: 108 Solution: The desired probability is the number of ways to pick the two isolated people times the probability that the remaining three point at each other. So, ! 3 3 2 5 1 5 2 P = · = 10 · = 4 2 6 108 2 is the desired probability.