HMMT 二月 2015 · COMB 赛 · 第 22 题
HMMT February 2015 — COMB Round — Problem 22
题目详情
HMMT February 2015 Saturday 21 February 2015 Combinatorics
- Evan’s analog clock displays the time 12:13; the number of seconds is not shown. After 10 seconds elapse, it is still 12:13. What is the expected number of seconds until 12:14?
- Victor has a drawer with 6 socks of 3 different types: 2 complex socks, 2 synthetic socks, and 2 trigonometric socks. He repeatedly draws 2 socks at a time from the drawer at random, and stops if the socks are of the same type. However, Victor is “synthetic-complex type-blind”, so he also stops if he sees a synthetic and a complex sock. What is the probability that Victor stops with 2 socks of the same type? Assume Victor returns both socks to the drawer after each step.
- Starting with the number 0, Casey performs an infinite sequence of moves as follows: he chooses a 1 number from { 1 , 2 } at random (each with probability ) and adds it to the current number. Let p m 2 be the probability that Casey ever reaches the number m . Find p − p . 20 15
- Alice Czarina is bored and is playing a game with a pile of rocks. The pile initially contains 2015 rocks. At each round, if the pile has N rocks, she removes k of them, where 1 ≤ k ≤ N , with each possible k having equal probability. Alice Czarina continues until there are no more rocks in the pile. Let p be the probability that the number of rocks left in the pile after each round is a multiple of 5. If p is of c a b the form 5 · 31 · , where a, b are integers and c, d are positive integers relatively prime to 5 · 31, find d a + b .
- For positive integers x , let g ( x ) be the number of blocks of consecutive 1’s in the binary expansion of x . For example, g (19) = 2 because 19 = 10011 has a block of one 1 at the beginning and a block 2 of two 1’s at the end, and g (7) = 1 because 7 = 111 only has a single block of three 1’s. Compute 2 g (1) + g (2) + g (3) + · · · + g (256).
- Count the number of functions f : Z → { ‘green’ , ‘blue’ } such that f ( x ) = f ( x + 22) for all integers x and there does not exist an integer y with f ( y ) = f ( y + 2) = ‘green’. 1
- 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?
- Let S be the set of all 3-digit numbers with all digits in the set { 1 , 2 , 3 , 4 , 5 , 6 , 7 } (so in particular, all three digits are nonzero ). For how many elements abc of S is it true that at least one of the (not necessarily distinct) “digit cycles” abc, bca, cab is divisible by 7? (Here, abc denotes the number whose base 10 digits are a , b , and c in that order.)
- Calvin has a bag containing 50 red balls, 50 blue balls, and 30 yellow balls. Given that after pulling out 65 balls at random (without replacement), he has pulled out 5 more red balls than blue balls, what is the probability that the next ball he pulls out is red?
- A group of friends, numbered 1 , 2 , 3 , . . . , 16, take turns picking random numbers. Person 1 picks a number uniformly (at random) in [0 , 1], then person 2 picks a number uniformly (at random) in [0 , 2], and so on, with person k picking a number uniformly (at random) in [0 , k ]. What is the probability that the 16 numbers picked are strictly increasing?
解析
22 . 5 45 9 the answer is = = . 65 130 26 ( )( )( ) 50 50 30 Solution 2. Let w ( b ) = be the number of possibilities in which b blue balls have b r = b +5 60 − 2 b been drawn (precisely 15 ≤ b ≤ 30 are possible). For fixed b , the probability of drawing red next is 50 − r 45 − b = . So we want to evaluate 50+50+30 − 65 65 ∑ 30 45 − b w ( b ) b =15 65 . ∑ 30 w ( b ) b =15 Combinatorics Note the symmetry of weights: ( )( )( ) ( )( )( ) 50 50 30 50 50 30 w (45 − b ) = = , 45 − b 50 − b 2 b − 30 b + 5 b 60 − 2 b 45 − (45 − b ) 45 / 2 45 − b 9 so the averages out with to give a final answer of = . 65 65 65 26 Remark. If one looks closely enough, the two approaches are not so different. The second solution may be more conceptually/symmetrically phrased in terms of the number of yellow balls.