HMMT 十一月 2020 · GEN 赛 · 第 4 题

HMMT November 2020 — GEN Round — Problem 4

Nine fair coins are flipped independently and placed in the cells of a 3 by 3 square grid. Let p be the probability that no row has all its coins showing heads and no column has all its coins showing tails. a If p = for relatively prime positive integers a and b , compute 100 a + b . b

解析

Nine fair coins are flipped independently and placed in the cells of a 3 by 3 square grid. Let p be the probability that no row has all its coins showing heads and no column has all its coins showing tails. a If p = for relatively prime positive integers a and b , compute 100 a + b . b Proposed by: Daniel Zhu Answer: 8956 Solution: Consider the probability of the complement. It is impossible for some row to have all heads and some column to have tails, since every row intersects every column. Let q be the probability that some row has all heads. By symmetry, q is also the probability that some column has all tails. We can then conclude that p = 1 − 2 q . 7 The probability that a given row does not have all heads is . So, the probability that none of the 8 ( ) 3 7 343 169 169 87 three rows have all heads is , implying that q = 1 − = . Thus p = 1 − = . 8 512 512 256 256