Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussels sprouts. Of these students, ~$\frac{2}{3}$~dislike lima beans; and of those who dislike lima beans, ~$\frac{3}{5}$~also dislike brussels sprouts. How many of the students like brussels sprouts but dislike lima beans?
(1) 120 students eat in the cafeteria.
(2) 40 of the students like lima beans.
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
所以题目中喜欢B的人不喜欢L=不喜欢L的人喜欢B
所以只要知道总人数就可以求,因此D
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B........| ? | 2/5 (题目求它)
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dis-B | ? | 3/5
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total | 1/3 | 2/3
不行哦 不给条件是只知道占比,题目问的是how many登录 或 注册 后可以参加讨论
https://www.beatthegmat.com/gmatprep-lima-beans-and-brussels-sprouts-t18453.html
B选项不过是用1/3先倒算出120人而已登录 或 注册 后可以参加讨论
2/3 dislike lima beans, 可以求到两类的人数;
再由3/5 also dislike brussels sprouts,求到一类的人数
所求值=2/3*2/5*总人数
哈喽 小晓 又见面了额
哈哈哈这是我之前刷题做的笔记😂登录 或 注册 后可以参加讨论
所有在这个餐厅吃饭的同学对LB和BS两种食物有单一偏好,所求内容为喜欢BS且不喜欢LB的学生人数,由题意,设总人数为N,2/3N的学生不喜欢LB,2/3*3/5N的学生不喜欢LB和BS,那么可以推出有(1-3/5)*2/3N的同学喜欢BS且不喜欢LB,所以只要确定N值即可,两个条件都分别单独能够确定N值,选D.
题目是问你P(A+B) = P(A) + P(B) - P(AB)
1. P(AB)=0是不充分的.因为如果同时发生的概率为0那么说明白球上都不是偶数,但是不代表不是偶数的都是白球.所以不充分
所以锁定选项B/C/E
2. 白球的概率-偶数概率为0.2,P(A) - P(B) = 0.2 .无法解出 P(A) + P(B)
所以两个加起来也不充分