请教保险公司记录正确率的问题

谢谢！

公司帐目用纸也用电子,纸不正确的公司里75%的电子不正确,电子不正确的公司里有60%的纸也不正确.总的公司里有3%两种都不正确,问随机抽取,两种都正确的概率

如果是这样的话，我算出来应该是6%

老jj

以前的解析给你参考

406PS:说:一项东东,即有纸报告,又有电子报告.纸报告有错的,60%电子也错;电子报告有错的,75%纸报告也错.发现共3%的纸报告和电子报告都错,问随便拿一个报告,都对的概绿.答案是:97,94,68,38,?. 反正94和68之间没数字.我选94,觉得错了,但不想再动脑筋想了.你们帮我想想看.以上数字不一定精确,但不差大数.	【答案】94%
	【思路】
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?	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%
Answer:94% E?
Bitmap
	题目中要求的是黄色的部分，即B。
	已知A+B+C+D+E+F=1，而现在E等于（D+E）的60%，也等于（E+F）的75%，但是算不出来A和C，所以B占的比例无法求出
	E/(E+D)=60%; E/(E+F)=75%
Bitmap	75% P’=3%X      ’=3%X/75%
	60%.E’=3%.X      E’=3%X/60%
	P& E=x-(3%X/60%+3%X/75%-3%X)
	纸错电错 ０.０３
	纸对电对 x
	纸错电对 y
	纸对电错 z
	(0.03+y)*75%=0.03
Bitmap	y=0.01
	(0.03+z)*60%=0.03
	z=0.02
	x=1-0.01-0.02-0.03=0.94
	(0.03+y)*75%=0.03是什么意思？
	在纸记录不正确的公司中，有75％电子记录也不正确
	即等于纸记录和电子记录都不正确３%
	~~~~~~~~~~~~~~~`
	what is the probability that a policy isn't recorded incorrectly? (~P & ~E)
	what is the probability that it is one having correct paper [or] correct electric record, [or both]? (P | E)
	As for the question "what is the probability that it is one having both correct paper and correct electric record?" (P & E), agree with 耳朵gg that the problem doesn't offer enough information to solve it. However, if the problem states that each policy has a paper record and an electric record, the question is solvable... a revised #1:
	In an insurance company, each policy has a paper record and an electric record. 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?
	60% P非=3%X
	75% E 非=3%X
	1-P非-E非+P非*E非

我搞错了，应该是94%，