HMMT 十一月 2012 · 冲刺赛 · 第 30 题
HMMT November 2012 — Guts Round — Problem 30
题目详情
- [ 15 ] A monkey forms a string of letters by repeatedly choosing one of the letters a , b , or c to type atrandom. Find the probability that he first types the string aaa before he first types the string abc .
HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND
英文原题
[ 15 ] A monkey forms a string of letters by repeatedly choosing one of the letters a , b , or c to type at
random. Find the probability that he first types the string aaa before he first types the string abc .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
HMMT NOVEMBER 2012, 10 NOVEMBER 2012 — GUTS ROUND
解析
英文解析
- [ 15 ]
Answer: It suffices to assume that the monkey starts all over as soon as he has typed a string 3
that ends in no prefix of either abc or aaa . For instance, if the monkey gets to abb we can throw these 7
out because there’s no way to finish one of those strings from this without starting all over.
Now, we draw the tree of all possible intermediate stages under this assumption; there are not manypossibilities. The paths from the root ”a” are:
a- aa- aaaa- aa- aab- aabca- ab- abc
The first and last possibilities have probability 1 / 27 each and the middle one has probability 1 / 81, so
1 / 27
in total the probability of getting the first before the second or the third is = .3
1 / 27+1 / 81+1 / 27 7
Guts Round