PUMaC 2010 · 组合（A 组） · 第 1 题
PUMaC 2010 — Combinatorics (Division A) — Problem 1
题目详情
- PUMaCDonalds, a newly-opened fast food restaurant, has 5 menu items. If the first 4 cus- tomers each choose one menu item at random, the probability that the 4th customer orders a previously unordered item is m/n , where m and n are relatively prime positive integers. Find m + n .
解析
- PUMaCDonalds, a newly-opened fast food restaurant, has 5 menu items. If the first 4 cus- tomers each choose one menu item at random, the probability that the 4th customer orders a previously unordered item is m/n , where m and n are relatively prime positive integers. Find m + n . Answer: 189 Solution: Number the menu items 1 through 5. Without loss of generality, assume the 4th customer orders menu item 1. Then the desired probability is the probability that each of 3 the first 3 customers do not order menu item 1, which is (4 / 5) = 64 / 125. The answer is 64 + 125 = 189 .