HMMT 二月 2014 · COMB 赛 · 第 3 题

HMMT February 2014 — COMB Round — Problem 3

Bob writes a random string of 5 letters, where each letter is either A , B , C , or D . The letter in each position is independently chosen, and each of the letters A, B, C, D is chosen with equal probability. Given that there are at least two A ’s in the string, find the probability that there are at least three A ’s in the string.

解析

Bob writes a random string of 5 letters, where each letter is either A , B , C , or D . The letter in each position is independently chosen, and each of the letters A, B, C, D is chosen with equal probability. Given that there are at least two A ’s in the string, find the probability that there are at least three A ’s in the string. ( ) ( ) 5 5 53 3 2 Answer: There are 3 = 270 strings with 2 A ’s. There are 3 = 90 strings with 3 A ’s. 188 2 3 ( ) ( ) 5 5 1 0 There are 3 = 15 strings with 4 A ’s. There is 3 = 1 string with 5 A ’s. 4 5 90+15+1 53 The desired probability is = . 270+90+15+1 188