HMMT 二月 2023 · 冲刺赛 · 第 6 题

HMMT February 2023 — Guts Round — Problem 6

[11] Let A , E , H , L , T , and V be chosen independently and at random from the set 0 , , 1 . Compute 2 the probability that b T · H · E c = L · A · V · A .

解析

[11] Let A , E , H , L , T , and V be chosen independently and at random from the set 0 , , 1 . Compute 2 the probability that b T · H · E c = L · A · V · A . Proposed by: Luke Robitaille 55 Answer: 81 3 3 Solution: There are 3 − 2 = 19 ways to choose L , A , and V such that L · A · V · A = 0, since at least 3 one of { L, A, V } must be 0, and 3 − 1 = 26 ways to choose T , H , and E such that b T · H · E c = 0, since at least one of { T, H, E } must not be 1, for a total of 19 · 26 = 494 ways. There is only one way to make b T · H · E c = L · A · V · A = 1, namely setting every variable equal to 1, so there are 495 total 6 55 ways that work out of a possible 3 = 729, for a probability of . 81