
求助两道新GRE的数学问题!!各位大侠求指教~!
1Acertainfive-menbercommitteemustbeassemblefromapoolof5women--ABCDandE,and3men--XYand...
1 A certain five-menber committee must be assemble from a pool of 5 women--A B C D and E, and 3 men--X Y and Z. What is the probability that the committee will include B C E Y and Z?
A 1/30 B 1/25 C 2/25 D 1/15 E 2/32
2 In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 people are in the club and 20 are neither on the team or the club, what is the minimum number of students who could, be both on the team and in the club?
数学不好。。。希望有详细过程!万分感谢~ 展开
A 1/30 B 1/25 C 2/25 D 1/15 E 2/32
2 In a group of 100 students, more students are on the fencing team than are members of the French club. If 70 people are in the club and 20 are neither on the team or the club, what is the minimum number of students who could, be both on the team and in the club?
数学不好。。。希望有详细过程!万分感谢~ 展开
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A
结果是选了三个女生,两个男生,
五个女生选三个,C(5,3)=10,三个男生选两个C(3,2)=3; 同时发生,所以相乘得30
其中一种情况概率1/30
61
100人中有20个两边都不是,所以至少参加一边的有80人
其中有70人确定是club的,那剩下10一定是team的
team那边又得比club多,所以至少得加61个人才能超过70
这61个人是从club那70人里出来的,所以
这61个人就是又在team又在club的最少值
结果是选了三个女生,两个男生,
五个女生选三个,C(5,3)=10,三个男生选两个C(3,2)=3; 同时发生,所以相乘得30
其中一种情况概率1/30
61
100人中有20个两边都不是,所以至少参加一边的有80人
其中有70人确定是club的,那剩下10一定是team的
team那边又得比club多,所以至少得加61个人才能超过70
这61个人是从club那70人里出来的,所以
这61个人就是又在team又在club的最少值
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