PUMaC 2015 · 团队赛 · 第 4 题

PUMaC 2015 — Team Round — Problem 4

[ 3 ] Ryan is messing with Brice’s coin. He weights the coin such that it comes up on one side twice as frequently as the other, and he chooses whether to weight heads or tails more with equal probability. Brice flips his modified coin twice and it lands up heads both times. The p probability that the coin lands up heads on the next flip can be expressed in the form for q positive integers p, q satisfying gcd( p, q ) = 1, what is p + q ?

解析

[ 3 ] Ryan is messing with Brice’s coin. He weights the coin such that it comes up on one side twice as frequently as the other, and he chooses whether to weight heads or tails more with equal probability. Brice flips his modified coin twice and it lands up heads both times. The p probability that the coin lands up heads on the next flip can be expressed in the form for q positive integers p, q satisfying gcd( p, q ) = 1, what is p + q ? Solution: Consider the entire space, where the coin could be weighted either way. Then, the 1 2 2 4 probability that the coin is weighted towards heads and the two flips are heads is · · = . 2 3 3 18 Similarly, the probability that the coin is weighted towards tails and the two flips are heads is 1 1 1 1 4 · · = . Then, after these first two flips, there is a probability that the coin is weighted 2 3 3 18 5 1 towards heads and towards tails. 5 4 2 1 1 9 3 On the next flip, the probability that the coin lands heads is then · + · = = , so 5 3 5 3 15 5 p + q = 8 . Author: Bill Huang 1