【MD每日一问】LR 1#

玛莎丁丁 · 发表于 2012-4-3 11:27:09

When 100 people who have not used cocaine are tested for cocaine use, on average only 5 will test positive. By contrast, of every 100 people who have used cocaine 99 will test positive. Thus, when a randomly chosen group of people is tested for cocaine use, the vast majority of those who test positive will be people who the have used cocaine.
A reasoning error in the argument is that the argument:
(B) attributes to every member of the population the properties of the average member of the population
(C)fails to take into account what proportion of the population have used cocaine

I chose B cause i thought the percentage calculated from mass research could not be applied to individual properly. Anyway, the key shows that C is the best choice. Could you provide me some help regarding to the mistake i made? Thanks.

simonshen · 发表于 2012-4-3 15:06:11

这题举个量化的例子比较好说明。假设一个randomly chosen的group里总共有1百万人，其中有100个是用过cocaine的，另外999,900都是没用过cocaine的。那按照平均数来讲，这100个用过cocaine的人里有99个会被测定为positive，而这999,900个没用过caocaine的人里，则会有999,000*0.05=49950个被test positive.所以这个group里被test positive的人的绝大多数其实是从未用过cocaine的。为什么会出现这个情况呢？因为这个group里用过cocaine跟没用过cocaine的人的比例严重失调，是1:9999，so this brings us to choice C.

玛莎丁丁 · 发表于 2012-4-3 15:15:52

got it! thank you=)

【MD每日一问】LR 1#

所属分类: 法学院申请

正在浏览此版块的会员 ()