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求教 TT GWD-12-6

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.

The given answer is D.  However I cannot  use condition 2 to get any integer value solution.

From condition 2,  I got  (n/10) x (10-n)/9 =  7/15  ====>  n x n - 10n + 42 = 0  ===> no integer solution

Did I get anything wrong?

Thanks!

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你的方程错了,应该是[(n) x (10-n)]*2/(9*10)=7/15。你少乘了一个2。

TOP

这不是有两个解么: 3和7, 不符合唯一性原则啊。。。。

TOP

题目中说A box contains 10 light bulbs, fewer than half of which are defective.

所以已经限定了 defective light bulbs <5

所以答案3唯一

TOP

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