OG12 CR78不解啊，望高人赐教！

anilolive

A recent report determined thatalthough only 3percent of drivers on Maryland highways equipped their vehicleswith radar detectors, 33 percent of all vehicles ticketed for exceeding thespeed limit were equipped with them. Clearly, drivers who equip theirvehicles with radar detectors are morelikely to exceed the speed limit regularly than are drivers who do not.
The conclusion drawn above depends on whichof the following assumptions?
A)  Drivers who equip their vehicles with radar detectorsare less likely to be ticketed for exceeding the speed limit than are driverswho do not.
B) Drivers who are ticketed for exceedingthe speed limit are more likely to exceed the speed limit regularly than aredrivers who are not ticketed.
C) The number of vehicles that were ticketed for exceeding the speed limitwas greater than the number of vehicles that were equipped with radardetectors.
D) Many of the vehicles that were ticketed for exceeding the speed limitwere ticketed more than once in the time period covered by the report.
E) Drivers on Maryland highways exceeded the speed limit more often thandid drivers on other state highways not covered in the report.

B是正确答案，但是这道题从题目到答案都让我觉得莫名其妙，望高人给个详解。不胜感激！

hoyancyn

刚开始看逻辑，LZ这道题我也看不懂啊，顶一个，楼下解答！

ItalyPirlo

題目 :  3%的drivers有裝rader detectors，但收到ticket中的drivers有33%有裝rader detectors 結論：裝rader detectors的drivers較沒裝rader detectors的drivers易超速。  

題目只給drivers裝rader detectors和收到ticket的關係，沒有給"超速"的關係，所以題目得出此結論是假設了(B) 收到ticket的drivers較沒收到tickets的drivers易超速。

如果那些收到tickets的drivers只是運氣背，被抓到超速，那就得不出裝rader detectors的drivers較沒裝rader detectors的drivers易超速。 只是那些被抓到超速的人正好大部分都有裝rader detectors而已。

以上是個人淺見。希望有幫道你，也希望高人賜教。

lavenderwu

这道题我也想了好久，终于想明白了。
关键就是超速不等于开罚单，也许1000人超速，只有100人收到罚单。这样这个假设就不成立。所以我们得限制一个前提就是：收到罚单的人确实比没收到罚单的人易超速，这样就否定了一种可能：没收到罚单的人也超速了。

togetherus

同意楼上，但是这题BACKGROUND 的装电子狗就没有用了么？是迷惑么？？？？？

zhanghaolin

这道题我想了好久，有两个诡异的做法：一是逻辑；用逻辑理解不了可以用数学方法做。

1、逻辑：结论与条件有gap，所以要填。条件是“装radar”和“所有vehicle”，结论是“装radar”和“ticketed”，因此选项的gap应该包含“所有vehicle”与“ticketed”；

2、数学：
条件(1)：装radar/所有=3%
条件(2)：装radar/ticketed=33%
结论：装>不装

(2)/(1)=所有/ticketed=11%

因此ticketed>所有

findjuhl

老兄，数学做法不对哦
条件1和条件2里面的 装radar 是不同的，不能拿来相除哦，他们还被你约了
比如共1000车，9车被ticketed，那么1中的 ‘装radar’ 是30车（3%*1000），2中的 ‘装radar’ 是3车（33%*9）

hoyancyn

还是挺疑问的，题目中提到了radar 和ticket的关系，难道是迷惑？

saffsa761

这题的主要原因是中国人的思维和老美的不同，我们已经习惯于某些假设认为是已知的了，所以想不出来