HMMT 二月 2004 · COMB 赛 · 第 8 题
HMMT February 2004 — COMB Round — Problem 8
题目详情
- Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 black balls. An urn is randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability that the first ball drawn was red, given that the second ball drawn was black?
解析
- Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 black balls. An urn is randomly selected, and then a ball inside of that urn is removed. We then repeat the process of selecting an urn and drawing out a ball, without returning the first ball. What is the probability that the first ball drawn was red, given that the second ball drawn was black? Solution: 7 / 15 This is a case of conditional probability; the answer is the probability that the first ball is red and the second ball is black, divided by the probability that the second ball is black. First, we compute the numerator. If the first ball is drawn from Urn A, we have a probability of 2 / 6 of getting a red ball, then a probability of 1 / 2 of drawing the second ball from Urn B, and a further probability of 3 / 6 of drawing a black ball. If the first ball is drawn from Urn B, we have probability 3 / 6 of getting a red ball, then 1 / 2 of drawing the second ball from Urn B, and 3 / 5 of getting a black ball. So our numerator is ( ) 1 2 1 3 3 1 3 7 · · + · · = . 2 6 2 6 6 2 5 60 We similarly compute the denominator: if the first ball is drawn from Urn A, we have a probability of 1 / 2 of drawing the second ball from Urn B, and 3 / 6 of drawing a black ball. If the first ball is drawn from Urn B, then we have probability 3 / 6 that it is red, in which case the second ball will be black with probability (1 / 2) · (3 / 5), and probability 3 / 6 that the first ball is black, in which case the second is black with probability (1 / 2) · (2 / 5). So overall, our denominator is ( [ ]) 1 1 3 3 1 3 1 2 1 · + · + · = . 2 2 6 6 2 5 2 5 4 Thus, the desired conditional probability is (7 / 60) / (1 / 4) = 7 / 15.