机经 145 超难求讨论

ceciliated

答案4/20中分母是怎么算的呢？求解释~

ceciliated

一个机构检查某总量为N进口产品, 发现2N/3（不是N/3）的产品里最多有3个不合格,2N/5的产品中最多有2个不合格, N/5的产品中最多有1个（不是2个）不合格, 求在全部N个产品中刚有2个不合格产品的概率. 选项：1/15,2/15,4/15，这三个确定，因为在这三个里面蒙的，另外两个貌似是1/3,1/5

A=2N/3: 最多有3个不合格: 0,1,2,3
B=N/5: 最多有1个: 0, 1
C=2N/15: 最多有1个: 0, 1

2+0+0; 1+1+0; 1+0+1; 0+1+1; 4 possibilities to get 2 bad products in (A+B+C) total products. Total possibities =5*2*2=20
Probability = 4/20 =1/5.

-- by 会员 sdcar2010 (2011/4/6 21:12:28)

should it be 4*2*2?

-- by 会员 blisserhk13 (2011/4/7 1:26:21)

答案4/20中分母的20是什么算的呢？求解释

sdcar2010

4/16 now.

thebeekeeper

再次翻出来问
另外楼上的
C=2N/15: 最多有1个: 0, 1，where do you get 2N/15?
4/16也没这个选项啊

Poodle1223

会不会是：

抽到<= 3个的概率是 ：2/3 =10/15；
抽到<=2个的概率是：2/5=6/15
抽到<=1个的概率是：1/5=3/15

<=2只有 0，1，2三种 ，<=1只有0，1两种，所以还是后两者相减？

那么第一句话要来干嘛呢。。。

sdcar2010

2N/3: 0,1,2,3 不合格
N/5: 0,1 不合格
2N/15:0,1 不合格

全部n个产品里，刚好有2个不合格: 0+1+1; 1+1+0; 1+0+1; 2+0+0; total 4 possibilities
Total possibilities = 4*2*2=16 possibilities
Probability = 4/16 =1/4

sdcar2010

The key is to figure out how to calculate the maximal number of bad products in (2N/15) products.

We know that every N/5 products have no more than 1 bad product. And we know that 2N/5, no more than 2. And 2N/3 , no more than 3. And 4N/5, no more than 4.

Therefore, (4N/5 - 2N/3) = 2N/15 will not have more than (4-3) = 1 bad product.

That's why I put 0, 1 for (2N/15) products!

2/3 + 1/5 + 2/15 = 1

sdcar2010

For this question, I prefer to have A has not more than 3, B has no more than 2, c has no more than 1, and A+B+C =N.  Then the things would become quite easy. Unfortunately 2/3 + 2/5 >1. So we cannot simply plug in those numbers.

Then the question is how to get the N exactly!

Basically 2N/5 and N/5 talk about the same thing. You just treat these two conditions as the same.

Then you have to plug in the 2N/3 condition. Now we are limited to ONLY 2N/3 and N/5. The difference between them and N is 2N/15. We need to know what is the POSSIBLE maximum number of products that are bad in 2N/15 products! One (1) is the answer as I just deduced above. It cannot be more than 1.

As to why not use ONLY 2N/5 and N/5 conditions, the reason is that you are omitting the 2N/3 condition by limiting yourself to 2N/5 and N/5!

sdcar2010

So the take home message is the following:

Figure out how to have A has no more than a bad products, B has no more than b bad products, C has no more than c bad products, and A+B+C =N. Most likely you will have to calculate the last condition from the combination of the first two.

It's rather simple if you follow this route

carvinhuang

So the take home message is the following:

Figure out how to have A has no more than a bad products, B has no more than b bad products, C has no more than c bad products, and A+B+C =N. Most likely you will have to calculate the last condition from the combination of the first two.

It's rather simple if you follow this route

-- by 会员 sdcar2010 (2011/4/10 12:40:00)

  请问~~~这道题，。。。能用中文解释一下吗~~~拜托拜托~~~看英语头好晕啊~~