LSAT-14-2-24 【airport scanner】

pocahontas

24. A certain airport security scanner designed to detect explosives in luggage will alert the scanner's operator whenever the piece of luggage passing under the scanner contains an explosive. The scanner will erroneously alert the operator for only one percent of the pieces of luggage that contain no explosives. Thus in ninety-nine out of a hundred alerts explosives will actually by present.

The reasoning in the argument is flawed because the argument

(A) ignores the possibility of the scanner's failing to signal an alert when the luggage does contain an explosive

(B) draws a general conclusion about reliability on the basis of a sample that is likely to be biased

(C) ignores the possibility of human error on the part of the scanner's operator once the scanner has alerted him or her

(D) fails to acknowledge the possibility that the scanner will not be equally sensitive to all kinds of explosives

(E) substitutes one group for a different group in the statement of a percentage

key's e. 是怎么想的？a和c为什么不行？

davecools

它说的是will erroneously alert the operator for only one percent of the pieces of luggage that contain no explosives

按这样分，total lugguage= luggage containing explosives(E) + luggage containing no explosives(NE)

1%=false report/NE

而Thus in ninety-nine out of a hundred alerts explosives will actually by present.

99%=false report/E

显然不能由100%-1%=99%这样推出上述结论，实际上错把对照组调换了。

lawyer_1

原文推理将erroneously alert the operator for only one percent of the pieces of luggage that contain no explosives理解为erroneously alert the operator for only one percent of the pieces of luggage （including luggage with and without explosives），即报警的错误率为1%，那末还有99%报警对的，即100个报警中有99个有EXPLOSIVES

pocahontas

got it! thax!!

Bensontuo

pocahontas 发表于 2004-7-29 08:52
24. A certain airport security scanner designed to detect explosives in luggage will alert the scann ...

Spot the question: Method of reasoning - Flaw.

Core of the argument

Total luggage = A+ B

1/100 of A = false alert. So 99/100 B must be true alert.

Apparently, A and B are switched by the argument.

Let us dive into answers.

A. It does not ignore, the methodology of calculating the possibilities is wrong.

B. The sample is not biased; instead, the methodology is biased.

C. Its away the core argument as we are discussing the "possibility " calculated by the operator, but not the possibilities of human errors.

D. Its not the " flaw " argument committed.

E. Totally matches what we just analysed