三道真题,一道不会,一道求证,一道求简便解法!!高人请进

pychina · 发表于 2007-12-22 05:06:00

If ab = 1, what is the value of (axb)(ayb)?
(1) ax = by = 2
(2) 2xy = 4

答案是D,这里ab,ax和by应该是个连在一起不可交换的,因为如果可以交换的话1和2将得出不同的结果,而且此题有类似的中文数学题规定是不可交换的,问题是怎么解呢,实在搞不明白,请高人指点!

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

答案是D,我选A,觉得2可能有两个解,不知各位怎么看.

Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5 ? 2/5 此题除了分情况讨论,各位还有什么更好的办法么,还是觉得只能分情况讨论也呢)

[此贴子已经被作者于2007-12-22 14:26:48编辑过]

yankevi · 发表于 2007-12-22 08:34:00

怎么是2/5啊？不是1/3吗？

pychina · 发表于 2007-12-22 11:48:00

因为可以变换顺序即26 ,62啊

pychina · 发表于 2007-12-24 02:35:00

顶上去,请大家看看第2题

pychina · 发表于 2007-12-24 02:35:00

顶上去,请大家看看第2题

home83 · 发表于 2007-12-24 03:01:00

不可能两个解

看清楚条件 fewer than half of which are defective

SO N<5 只有一个解 3 7是不可以的

manfromars · 发表于 2007-12-24 10:14:00

第一题不要想太复杂吧，ets就是有这么多简单题目。想复杂了这题就无解了。并没要求说算出来一样，只要能求解就行了。支持D

第二题，条件2我求出来无解
(10-n)/10 * n/9=7/15 => (n-5)^2+17=0
无解
选A

第三题，穷举(26,62,53,35,44)
2/5

龙飞凤舞 · 发表于 2007-12-30 06:37:00

以下是引用manfromars在2007-12-24 10:14:00的发言：
第一题不要想太复杂吧，ets就是有这么多简单题目。想复杂了这题就无解了。并没要求说算出来一样，只要能求解就行了。支持D

第二题，条件2我求出来无解
(10-n)/10 * n/9=7/15 => (n-5)^2+17=0
无解
选A

第三题，穷举(26,62,53,35,44)
2/5

第三题怎么会是2/5啊?

分母就是出现数字的可能, 不是相加吗? 那就应该最小1+1=2, 最大6+6=12, 中间都行, 那不就是2/11吗?

我想错了?

Taken · 发表于 2007-12-30 10:45:00

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

(1) C(2,n)/C(2,10)=1/15, n=3
(2) C(1,n)C(1,10-n)/C(2,10)=7/15, n=3 or n=7; since n<10/2<5, n=3

key

yangflyfree · 发表于 2007-12-30 22:35:00

我尽量用大家都看得懂的方式来表达。。。

第二题第二个数据分析

情况1：第一个坏的第二个好的

情况2：第一个好的第二个坏的

相加等于7/15

第三道题列举法很简单由于加起来等于8

所以无非五种情况：2，6 ；3，5；4，4；5，3；6，2。其中2种有数字5所以概率就是2/5

三道真题,一道不会,一道求证,一道求简便解法!!高人请进

三道真题,一道不会,一道求证,一道求简便解法!!高人请进

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