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标题: Can anyone help me out? thanks [打印本页]

作者: jessie_dai 时间: 2003-6-3 06:02 标题: Can anyone help me out? thanks

1.The Food and Drug Administration (FDA) reported that medication errors are categorized into three groups: minor, moderate and severe. The probability that a given medication error is minor is 0.5, that it is moderate is 0.4, and that it is severe is 0.1.
If two medication errors occur independently in one month, what is the probability that neither is severe and at most one is moderate.

作者: robert 时间: 2003-6-3 10:16

答:

根据题意可以得出，这两个errors有两种情况，一种是 (min,min)，还有一种是(min,mod)，
所以概率为 0.5*0.5+0.5*0.4=0.25+0.2=0.45

有不同看法的可以回帖讨论

作者: jessie_dai 时间: 2003-6-3 12:50

still confusing...

coz I feel it is more make sense to exclude the other possiblities :

1-(0.1*0.1+0.1*0.5+0.1*0.4+0.4*0.4)=0.74???

I also think of 0.45 as the anwer, but if using the above method, the possibility of sever 0.1 seems useless here?

作者: frances 时间: 2003-6-11 22:13

同意robert 。
jessie_dai：我是这样理解的：不能用1去减，因为无论是哪两个错同时发生，整体的概率都不是100%，也就不可能是“1”！不知你是否同意我的说法？

作者: fengqiwu 时间: 2003-7-3 15:09

同意Robert之说法，occur independently用乘法法则，两种情况，相加即可

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