GWD 5 2

jefferyli

Q2:

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?

(1)     More than 1/2 of the 10 employees are women.

(2)     The probability that both representatives selected will be men is less than 1/10.

                  

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.

(E)

怎样快速解答？花了6分钟的说。

fair_sword

me too,

背下来算了，

daisy9023305

请教此题思路如何计算？,谢谢

tasw76

(1)和(2)好像是等价的呀。男4女6，则选出同为女概率是1/3，男3女7则选出同为女是7/15，

男2女8则选出同为女概率是28/45 （这时p>1/2了）

tiger1234

you can just do what tasw76 did, it's probably easier, but here's how to think of it with formulas:

n chooses k = n! / (k! * (n-k)!), means how many ways can one choose k things from a set of n things(of course n >= k)

let w = the number of women;

then p = w chooses 2 / 10 chooses 2 (w chooses 2 gives how many ways u can choose any 2 women out of all the women, use that divide how many ways you can choose 2 any people out of the 10 people, that gives you the probability that both ppl selected are women);

do some calculations and you will find that p is simply w * (w-1) / 90;

if w > 5, then follow the example of tasw76, if w = 7, p = 6 * 7 / 90 = 42 / 90 < 1/2 ; but if w = 10, then p = 1 > 1/2, thus (1) by itself is not sufficient;

let q = m chooses 2 / 10 chooses 2, where m = the number of men; (this is the probability that both ppl selected are men)

and q = m * (m - 1) / 90;

if q < 1/10, that means m * (m - 1) / 9 < 1, so m could be 0, 1, 2, 3, let it be 0, that means w = 10 - 0 = 10, so p = 1 > 1/2; but if m = 3, then w = 10 - 3 = 7, and p = 42/90 < 1/2, so (2) is not sufficient by itself;

if you take these two together, you will notice the same thing happens with the example of 10 women and 0 man, and th example of 7 women and 3 men, thus (1) and (2) taken together is not sufficient;

Therefore, the answer is E;

Personally I'd just do what tasw76 did, since the set for this problem is not that big;

daisy9023305

thank you very much!!everybody~~

winnowerfly

我这样算法是否够严密呢?

设女的有X人.求Cx2/C102 >1/2

即X*(X-1)/10*9>1/2 即X*(X-1)>45,那X只能>=8了.

(1)X>5,不一定>=8,不sufficient

(2)同样算一下,设男的为Y,那么Y*(Y-1)/10*9<1/10就是Y<=3

等于3就不sufficient了.

zhaoyak7

以下是引用winnowerfly在2006-7-6 23:41:00的发言：

我这样算法是否够严密呢?

设女的有X人.求Cx2/C102 >1/2

即X*(X-1)/10*9>1/2 即X*(X-1)>45,那X只能>=8了.

(1)X>5,不一定>=8,不sufficient

(2)同样算一下,设男的为Y,那么Y*(Y-1)/10*9<1/10就是Y<=3

等于3就不sufficient了.

偶也是这个算法，应该为B。顶吧

cournot

b is wrong=>E is right

assume w is the number of women,then 10-w is the number of men

C(1,10-w)*C(1,9-w)/C(1,10)*C(1,9)<1/10

(10-w)*(9-w)<9

w=10,9,8,7

when w=7,p=C(1,7)*C(1,6)/C(1,10)*C(1,9)<1/2

when w=8,9,10,p>1/2

as for (1): w>5

p=C(1,w)*C(1,w-1)/C(1,10)*C(1,9)

when w=6 or 7,p<1/2

when w=8,9,10,p>1/2

so (1) and (2) together are not sufficient