以下是引用allen0018在2006-8-16 13:02:00的发言: 1、In an insurance company, each policy has a paper record or an electric record, or both of them. 60 percent of the policies having incorrect paper record have incorrect electric record and 75 percent of the policies having incorrect electric record have incorrect paper record. 3 percent of all the policies have both incorrect paper and incorrect electric records. If we randomly pick out one policy, what is the probability that it is one having both correct paper and correct electric record? 【答案】94% 【思路】 设总数为X policies,则既有非P又有非E为3%X;60%的非P中为非E,则非P为3%X/60%;75%的非E为非P,则非E为3%/75%,那么非P与非E一共有(3%X/60%+3%X/75%-3%X) ,那么P与E一共有X-(3%X/60%+3%X/75%-3%X),则其概率为[X-(3%X/60%+3%X/75%-3%X)]/X=94% 这是feifei 上面的,看的我简直要抓矿,谁方便能用人类的语言说说到底怎么回事 讨论帖也都是行为艺术  我来试试吧。正确的纸版记为P, 不正确的记为“非P”,正确的电子版记为E, 不正确的记为“非E" 由题意可知: 非P 非E : 非P E = 60% : 40% 非E 非P : 非E P = 75% : 25% 而题中给出了 非P 非E (也就是 非E非P)=3% 因此可以算出 非P=非P非E + 非P E = 非P非E/60%= 3%/60%=5% 非E=非E非P + 非E P = 非E非P/75%=3%/75%=4% 因此题中问到的PE都正确的概率 即:1 - 非P - 非E + 非P非E = 1 - 5% - 4% + 3% =94% 因为非P中包含了非P非E,非E中也包含了非P非E,因此要加回一个非P非E. |