Background information: This year, each film submitted to the Barbizon Film Festival was submitted in one of ten categories. For each category, there was a panel that decided which submitted films to accept.
Fact 1: Within each category, the rate of acceptance for domestic films was the same as that for foreign films.
Fact 2: The overall rate of acceptance of domestic films was significantly higher than that of foreign films.
In light of the background information, which of the following, if true, can account for fact 1 and fact 2 both being true of the submissions to this year’s Barbizon Film Festival?
这道题目给的参考答案是E,但是我觉得这道题目应该选B
这应该是一道数学逻辑题目,
题目中有两个事实:1,每个分类中,本地电影和外来电影的通过率是相同的;2,所有的加起来,本地电影的通过率远远大于外来电影。
问解释。
可以这么来看,为了简单期间认为就两个分类,且设第1个分类中本地电影总数为A1,外来电影总数为B1,相应的第2个分类分别为A2和B2。
那么从事实1可以得到通过率一样,此处设为x,为了简单期间认为两个分类通过率相同。
从事实2可以得到(A1X+A2X) / (A1+A2+B1+B2) > (B1X+B2X) / (A1+A2+B1+B2),同时消掉分母,即得到A1+A2 > B1+B2,也就是B答案所描述的内容。
请讨论。
可以这么来看,为了简单期间认为就两个分类,且设第1个分类中本地电影总数为A1,外来电影总数为B1,相应的第2个分类分别为A2和B2。
那么从事实1可以得到通过率一样,此处设为x,为了简单期间认为两个分类通过率相同。
从事实2可以得到(A1X+A2X) / (A1+A2+B1+B2) > (B1X+B2X) / (A1+A2+B1+B2),同时消掉分母,即得到A1+A2 > B1+B2,也就是B答案所描述的内容。
==============================
有一点问题
最后是 A1*x1+A2*x2+...+A10*x10>B1*x1+B2*x2+...+B10*x10
即 x1*(A1-B1)+x2*(A2-B2)+...+x10*(A10-B10)>0
不能推出 A总>B总
只能推出,B在那些通过率低的种类里可能数量上比在那些通过率高的种类里多一些。
B 从数量上来讲是可以解释的,但有缺陷
fact 1:其实是句废话,传达的意思就是国内国外都是一视同仁的
FACT 2:结果是国内比国外的多N多
逻辑:1.影片一视同仁-->panel如何选片呢?-->影片质量要好-->质量好的能入选
B:说国内影片数量多,国外影片数量少,则出现FACT2结果的情况就一种,由10个电影组成一组的category中,必须有N多这样的组是由全部的国内影片组成,从而在一致的录取率下,国内影片入围的几率才能比国外电影多N多。但是题目不严密,没有说每组必须有外国电影,所以这样的极端情况也是可能存在的。可是还有一种情况,要是国内电影虽然量多,但是品质不高也是不会录取的。所以这也是B的weakness。所以B有点问题,但也不全错
E:说了影片质量很好,但是伴随着很低的录取率,是说了另一个事实,没有说理由是什么,即是FACT2的另外一种说法而已。所以肯定是对的咯。因为它重复了一下废话。
相比E更好吧
我其实一开始也选B~~~~
汗一个
Most foreign films, unlike most domestic films, were submitted in categories with high prestige, but with correspondingly low rates of acceptance.
这句话的意思是 外语片多分在通过率低的种类里了
Most foreign films, unlike most domestic films, were submitted in categories with high prestige, but with correspondingly low rates of acceptance.
这句话的意思是 外语片多分在通过率低的种类里了
谢谢,理解了,我原来是把这句话理解错了。
A:无关
B:无关
C In each of the past three years无关
D predetermined但没提到具体值,所以无关
E f电影被送到高声望低接受率得组得组,因为没提到f和d总量大小,所以有三种情况(仅设两组,可以推广到10组得情况)
1)f=d
设f=100,d=100
普通组
D送了90,通过80%为72,f送10,通过80%,8
高声望低接受组
D送10,通过20%,2,f送90,20%,18
总体:d通过率=(72+2)/100=74%
F通过率=(8+18)/100=26%
2)f>d设f=200,d=100
普通组
D送了90,通过80%为72,f送10,通过80%,8
高声望低接受组
D送10,通过20%,2,f送190,20%,38
总体:d通过率=(72+2)/100=74%
F通过率=(8+38)/200=23%
3)f<d设f=100,d=200
普通组
D送了190,通过80%为152,f送10,通过80%,8
高声望低接受组
D送10,通过20%,2,f送90,20%,18
总体:d通过率=(152+2)/200=77%
F通过率=(8+18)/100=26%
A:无关
B:无关
C In each of the past three years无关
D predetermined但没提到具体值,所以无关
E f电影被送到高声望低接受率得组得组,因为没提到f和d总量大小,所以有三种情况(仅设两组,可以推广到10组得情况)
1)f=d
设f=100,d=100
普通组
D送了90,通过80%为72,f送10,通过80%,8
高声望低接受组
D送10,通过20%,2,f送90,20%,18
总体:d通过率=(72+2)/100=74%
F通过率=(8+18)/100=26%
2)f>d设f=200,d=100
普通组
D送了90,通过80%为72,f送10,通过80%,8
高声望低接受组
D送10,通过20%,2,f送190,20%,38
总体:d通过率=(72+2)/100=74%
F通过率=(8+38)/200=23%
3)f<d设f=100,d=200
普通组
D送了190,通过80%为152,f送10,通过80%,8
高声望低接受组
D送10,通过20%,2,f送90,20%,18
总体:d通过率=(152+2)/200=77%
F通过率=(8+18)/100=26%
顶,分析得精彩.
本题的关键在于对"each film submitted to the Barbizon Film Festival was submitted in one of ten categories."和"Within each category, the rate of acceptance for domestic films was the same as that for foreign films."的理解.
我一开始把前者理解为"该Festival上总共就十部电影,它们的种类各不相同,也就是每一种类一部",结果把后者理解为"D与F被接受的概率都是50%",最终导致选B,还认为很正确. 其实是曲解了题意............
可以这么来看,为了简单期间认为就两个分类,且设第1个分类中本地电影总数为A1,外来电影总数为B1,相应的第2个分类分别为A2和B2。
那么从事实1可以得到通过率一样,此处设为x,为了简单期间认为两个分类通过率相同。
从事实2可以得到(A1X+A2X) / (A1+A2+B1+B2) > (B1X+B2X) / (A1+A2+B1+B2),同时消掉分母,即得到A1+A2 > B1+B2,也就是B答案所描述的内容。
请讨论。
设通过率为X的时候应该考虑到每个种类的通过率是不同的,对于种类1和种类2应当对应通过率X1和X2。
这样推出来的式子就是A1X1+A2X2>B1X1+B2X2, 可以找到反例满足上式但不满足A1+A2>B1+B2
而我们这里设了每个种类不同的通过率也正是选项E的意思。
Most foreign films, unlike most domestic films, were submitted in categories with high prestige, but with correspondingly low rates of acceptance.
这句话的意思是 外语片多分在通过率低的种类里了
关键点是这里。。
天呀你们这么会想得这么复杂???!!!
假设外语片和国语片各有110部.共有2个categorys.
在high prestige那一组有100部外语片竞选, 只有10国语片竞选. 通过率只有10%: 入围外语片10部, 入围国语片1部.
在另一组只有10部外语片和100部国语片竞选,通过率只有50%: 入围外语片5部, 入围国语片50部.
这就说明了E是正确的.
楼上见解精辟
Fact 1: Within each category, the rate of acceptance for domestic films was the same as that for foreign films. Exante rate of acceptance for its sample
There is unbiased judgment.
Fact 2: The overall rate of acceptance of domestic films was significantly higher than that of foreign films. Postante rate of acceptance for its sample
The 1rst rate of acceptance compare between the accepted domestic films and accepted foreign films; the second rate of acceptance compare acceptance ratio between both category
Higher acceptance rate for domestic film and foreign film in the population
Most foreign films, unlike most domestic films, were submitted in categories with high prestige, but with correspondingly low rates of acceptance. (rate of acceptance in general in each category)
If more foreign films were submit to the categories of higher prestige, categories also have lower rate of acceptance. The overall rate of acceptance would be bring down by the films in such category.
Significantly more domestic films than foreign films were submitted to the festival. The overall rate of acceptance is not affected by the sample size but by rate of acceptance for each film in the sample. This evidence may lead to larger quantity of domestic films accepted this year, but such number is also scaled by the larger sample N.
欢迎光临 ChaseDream (https://forum.chasedream.com/) | Powered by Discuz! X3.3 |