LSAT-14-2-24 【airport scanner】

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？

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。