HMMT 二月 2023 · 冲刺赛 · 第 20 题

HMMT February 2023 — Guts Round — Problem 20

[16] Five people take a true-or-false test with five questions. Each person randomly guesses on every question. Given that, for each question, a majority of test-takers answered it correctly, let p be the a probability that every person answers exactly three questions correctly. Suppose that p = where a is b 2 an odd positive integer and b is a nonnegative integer. Compute 100 a + b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2023, February 18, 2023 — GUTS ROUND Organization Team Team ID# 2 2

解析

[16] Five people take a true-or-false test with five questions. Each person randomly guesses on every question. Given that, for each question, a majority of test-takers answered it correctly, let p be the a probability that every person answers exactly three questions correctly. Suppose that p = where a b 2 is an odd positive integer and b is a nonnegative integer. Compute 100 a + b . Proposed by: Evan Erickson, Leo Yao, Reagan Choi Answer: 25517 5 Solution: There are a total of 16 ways for the people to collectively ace the test. Consider groups of people who share the same problems that they got incorrect. We either have a group of 2 and a group of 3, or a group 5. ( ) ( ) 5 5 In the first case, we can pick the group of two in ways, the problems they got wrong in ways. 2 2 Then there are 3! ways for the problems of group 3. There are 600 cases here. In the second case, we can 5! · 4! / 2 = 120 · 12 ways to organize the five cycle (4! / 2 to pick a cycle and 5! ways to assign a problem to each edge in the cycle). 255 Thus, the solution is and the answer is 25517. 17 2 2 2