HMMT 二月 2004 · COMB 赛 · 第 8 题

英文解析

8. Urn A contains 4 white balls and 2 red balls. Urn B contains 3 red balls and 3 blackballs. An urn is randomly selected, and then a ball inside of that urn is removed. Wethen repeat the process of selecting an urn and drawing out a ball, without returningthe 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 firstball is red and the second ball is black, divided by the probability that the second ballis black.
   First, we compute the numerator. If the first ball is drawn from Urn A, we have aprobability of 2 / 6 of getting a red ball, then a probability of 1 / 2 of drawing the secondball from Urn B, and a further probability of 3 / 6 of drawing a black ball. If the firstball is drawn from Urn B, we have probability 3 / 6 of getting a red ball, then 1 / 2 ofdrawing the second ball from Urn B, and 3 / 5 of getting a black ball. So our numeratoris
   ( )
   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 ablack 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 withprobability (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.