今天一道数学题把我给难住了!

splendor

Each document has both a paper copy and an electrical copy. Of the paper copies that are inaccurate, 60% also has an inaccurate electrical copies. Of the electrical copies that are inaccurate, 75% also has an inaccurate paper copy. 3% of documents has both inaccurate paper copy and electric copy. What's the probability a randomly drawn document has both accurate paper and accurate electrical copy?

charmant

记E为事件Electrical copy错误,P为Paper copy错误

由题意知：P(P|E)=75%, P(E|P)=60%   (EP)=3%

根据条件概率公式：P(P|E)=P(EP)/P(E)

所以P(E)=P(EP)/P(P|E)=3%/75%=4%

        同理可得P(P)=5%

所以P(E+P)=P(P)+P(E)-P(EP)=6%

所以Both accurate的probability=1-6%=94%

用文氏图应该是

[attachimg]24246[/attachimg]

[此贴子已经被作者于2004-12-28 13:12:24编辑过]

blackhorse

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感谢你的理解和支持！

而且这道题目很多人问过了，建议你搜索以前的讨论。

iloveqi

以下是引用splendor在2004-12-28 11:49:00的发言：
Each document has both a paper copy and an electrical copy. Of the paper copies that are inaccurate, 60% also has an inaccurate electrical copies. Of the electrical copies that are inaccurate, 75% also has an inaccurate paper copy. 3% of documents has both inaccurate paper copy and electric copy. What's the probability a randomly drawn document has both accurate paper and accurate electrical copy?

好像本月几经里见过，inaccurateP&E占总的3%，又分别在inaccurateP和inaccurateE中占60%和75%，由此，可以求出inaccurateP占总的：3%除以60%=5%，inaccurateE占总的：3%除以75%=4%，所以accurateP&E=1-5%-4%+3%=94%

blackhorse

以下是引用iloveqi在2004-12-28 13:05:00的发言：


好像本月几经里见过，inaccurateP&E占总的3%，又分别在inaccurateP和inaccurateE中占60%和75%，由此，可以求出inaccurateP占总的：3%除以60%=5%，inaccurateE占总的：3%除以75%=4%，所以accurateP&E=1-5%-4%+3%=94%

数学不错，鼓励！！！

zhaoshujuanw

Thanks ,big NN

爱维的天

以下是引用blackhorse在2004-12-28 12:53:00的发言：

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感谢你的理解和支持！

而且这道题目很多人问过了，建议你搜索以前的讨论。

我推这个…

soaring

charmant真牛~

feichiwu

低等数学解法：设一共有100m，其中纸纪录错误的y个，电子纪录错误x个。75%x=3%*100m 所以，x=4m,同理，y=5m

没错的100m-5m-4m-3m=94m

94m\100m=94%