HMMT 二月 2007 · 冲刺赛 · 第 21 题
HMMT February 2007 — Guts Round — Problem 21
题目详情
- [ 10 ] Bob the bomb-defuser has stumbled upon an active bomb. He opens it up, and finds the red and green wires conveniently located for him to cut. Being a seasoned member of the bomb-squad, Bob quickly determines that it is the green wire that he should cut, and puts his wirecutters on the green wire. But just before he starts to cut, the bomb starts to count down, ticking every second. Each time the bomb ticks, starting at time t = 15 seconds, Bob panics and has a certain chance to move his 1 wirecutters to the other wire. However, he is a rational man even when panicking, and has a chance 2 2 t of switching wires at time t , regardless of which wire he is about to cut. When the bomb ticks at t = 1, Bob cuts whatever wire his wirecutters are on, without switching wires. What is the probability that Bob cuts the green wire? 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND
解析
- [ 10 ] Bob the bomb-defuser has stumbled upon an active bomb. He opens it up, and finds the red and green wires conveniently located for him to cut. Being a seasoned member of the bomb-squad, Bob quickly determines that it is the green wire that he should cut, and puts his wirecutters on the green wire. But just before he starts to cut, the bomb starts to count down, ticking every second. Each time the bomb ticks, starting at time t = 15 seconds, Bob panics and has a certain chance to move his 1 wirecutters to the other wire. However, he is a rational man even when panicking, and has a chance 2 2 t of switching wires at time t , regardless of which wire he is about to cut. When the bomb ticks at t = 1, Bob cuts whatever wire his wirecutters are on, without switching wires. What is the probability that Bob cuts the green wire? 23 Answer: . Suppose Bob makes n independent decisions, with probabilities of switching p , p , . . . , p . 1 2 n 30 Then in the expansion of the product P ( x ) = ( p + (1 − p ) x )( p + (1 − p ) x ) · · · ( p + (1 − p ) x ) , 1 1 2 2 n n the sum of the coefficients of even powers of x gives the probability that Bob makes his original decision. This is just ( P (1) + P ( − 1)) / 2 , so the probability is just ( ) ( ) ( ) 1 1 1 14 16 13 15 1 3 8 1 + 1 − 1 − · · · 1 − 1 + · · · 1 + 23 15 15 14 14 2 2 15 15 14 14 2 2 15 = = = . 2 2 2 30 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 10 HARVARD-MIT MATHEMATICS TOURNAMENT, 24 FEBRUARY 2007 — GUTS ROUND