HMMT 十一月 2016 · 冲刺赛 · 第 25 题
HMMT November 2016 — Guts Round — Problem 25
题目详情
英文原题
- [ 13 ] Chris and Paul each rent a different room of a hotel from rooms 1 − 60. However, the hotelmanager mistakes them for one person and gives ”Chris Paul” a room with Chris’s and Paul’s roomconcatenated. For example, if Chris had 15 and Paul had 9, ”Chris Paul” has 159. If there are 360
rooms in the hotel, what is the probability that ”Chris Paul” has a valid room?
解析
英文解析
- [ 13 ] Chris and Paul each rent a different room of a hotel from rooms 1 − 60. However, the hotelmanager mistakes them for one person and gives ”Chris Paul” a room with Chris’s and Paul’s roomconcatenated. For example, if Chris had 15 and Paul had 9, ”Chris Paul” has 159. If there are 360
rooms in the hotel, what is the probability that ”Chris Paul” has a valid room?
Proposed by: Meghal Gupta
Answer:153
1180
There are 60 · 59 = 3540 total possible outcomes, and we need to count the number of these whichconcatenate into a number at most 60. Of these, 9 · 8 result from both Chris and Paul getting one-digitroom numbers. If Chris gets a two-digit number, then he must get a number at most 35 and Paulshould get a one-digit room number, giving (35 − 9) · 9 possibilties. If Chris gets a one-digit number, itmust be 1, 2, or 3. If Chris gets 1, 2 or 3, Paul can get any two-digit number from 10 to 60 to guaranteea valid room, giving 51 · 3 outcomes. The total number of correct outcomes is 72 + 51 ∗ 3 + 26 ∗ 9 = 459,
so the desired probability is .153
1180