HMMT 二月 2015 · COMB 赛 · 第 7 题
HMMT February 2015 — COMB Round — Problem 7
题目详情
- 2015 people sit down at a restaurant. Each person orders a soup with probability . Independently, 2 1 each person orders a salad with probability . What is the probability that the number of people who 2 ordered a soup is exactly one more than the number of people who ordered a salad?
解析
- 2015 people sit down at a restaurant. Each person orders a soup with probability . Independently, 2 1 each person orders a salad with probability . What is the probability that the number of people who 2 ordered a soup is exactly one more than the number of people who ordered a salad? Combinatorics 4030 4030 4032 4030 − 2 ( ) ( ) ( ) ( ) 2016 2014 2016 2015 Answer: OR OR Solution 1. Note that total soups = total salads + 1 4030 4030 4031 2 2 2 ( ) 2015+2015 is equivalent to total soups + total not-salads = 2016. So there are precisely possibilities, 2016 4030 ( ) 2015+2015 2016 each occurring with probability (1 / 2) . Thus our answer is . 4030 2 ( )( ) ∑ 2014 2015 2015 Solution 2. To count the number of possibilities, we can directly evaluate the sum . i =0 i i +1 ( ) ( ) ( )( ) ∑ 2014 2015 2015 2015 2015 One way is to note = , and finish with Vandermonde’s identity: = i =0 i +1 2014 − i i 2014 − i ( ) ( ) ( ) 2015+2015 4030 4030 = (which also equals ). 2014 2014 2016 ( ) ( ) ( )( ) ( ) ∑ 2014 2015 2015 2015 2015 2015+2015 (We could have also used = to get = directly, which is i =0 i 2015 − i 2015 − i i +1 2016 closer in the spirit of the previous solution.) ( )( ) ∑ 2014 2015 2015 Solution 3 (sketch). It’s also possible to get a handle on by squaring Pascal’s i =0 i i +1 4032 4030 ( ) ( ) ( ) − 2 ( ) ( ) 2015 2015 2016 2016 2015 identity + = and summing over 0 ≤ i ≤ 2014. This gives an answer of , 4031 i i +1 i +1 2 ( ) ( ) 4032 4031 4032 which can be simplified by noting = , and then applying Pascal’s identity. 2016 2016 2015