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标题: GWD-3-6 请教 [打印本页]

作者: hedgeforfun    时间: 2005-11-7 07:15     标题: GWD-3-6 请教

Q6:

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?

(1) The probability that the two bulbs to be drawn will be defective is 1/15.

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

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.

标准答案是D,我的是A,因为由(2)推不出可能的正整数n:(n/10)((10-n)/9)=7/15???

请教那位可以帮忙解释我的错误?


作者: sammen    时间: 2005-11-7 10:03

I agree with u. from 2, we can know n may be 3 or 7.
作者: goodhope    时间: 2005-11-7 16:16

那就对了, 人家题中说是N小于 5的。 那就是3个了。


作者: asasasas1    时间: 2005-11-8 06:55

N*(10-N)/C(10,2)=7/15

N*(10-N)=21

N<5,

N=3

是同时抽,所以是C(10,2). 先后才是10,9


作者: hedgeforfun    时间: 2005-11-8 13:03

感谢!

想明白了,我的计算错了:就是先后抽,这道题也有解...(n/10)((10-n)/9)*2=7/15

so, n=3






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