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发表于 24-12-2013 12:56 AM | 显示全部楼层 |阅读模式
本帖最后由 grass83 于 23-10-2016 09:52 PM 编辑

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发表于 24-12-2013 01:13 AM | 显示全部楼层
因为问题已经说明,ELEVEN PIECES OF CARDS,所以会有重复的字母在box里。
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 楼主| 发表于 24-12-2013 01:21 AM | 显示全部楼层
尚杰 发表于 24-12-2013 01:13 AM
因为问题已经说明,ELEVEN PIECES OF CARDS,所以会有重复的字母在box里。

尚杰,谢谢 可是在Exp2 里,很明显是有5个vowels . 但 却n(F) = 3(因为有重复)。 同样是 n(F) 在probability里也算是set的写法,可是 n(F) 却不一样。
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 楼主| 发表于 24-12-2013 11:04 AM | 显示全部楼层
The semantics of the term set imposes certain syntactic constraints on these notations. The only information that is fundamental for a set is which particular objects are, or are not, elements. As a result, the order in which elements appear in an enumeration is irrelevant: {π, 6, 1/2} and {1/2, π, 6} are two enumerations of a single set. Likewise, repeated mention of an element is also irrelevant, so {1, 2, 2, 3, 3, 3} = {1, 2, 3}                                                              (Wikepedia)

Duplicate Elements in Mathematical Sets

Date: 08/31/2004 at 11:49:20
From: Derek
Subject: Duplicate Elements in Mathematical Sets

Can there be duplicate elements in a mathematical set?

I've read definitions which suggest that a set of [red, blue, green]
is the same set as [red, green, red, blue, blue, red].  Which suggests
that duplications do not matter.  Computer programming language sets
tend to be defined in such a way that duplicates are not allowed.  So
just what IS the true definition?

In event that duplicate elements are not allowed in a set, then are
1/2 and 2/4 considered duplicate elements?  In my mind, they are
uniquely indentifiable as being distinct elements, much like the words
"black" and "noir."  Technically speaking, a person who has no math
background (!) would state that 1/2 and 2/4 are obviously different.  
Much as an alien to our world with no understanding of our language
would never even think to equate "black" and "noir" even though they
in fact do refer to the same colour.

The intersection of the sets [1,2] and [2,3] is [2], correct?  But if
the sets were [1/2,1/3] and [2/4,3/9] then would the intersection be
an empty set or... what?


Date: 08/31/2004 at 14:22:40
From: Doctor Mike
Subject: Re: Duplicate Elements in Mathematical Sets

Hi Derek -

Thanks for writing to Dr. Math.  You've asked a variety of questions
which are all related, and hopefully when I've answered each of them,
the whole idea will make sense.

First, can there be duplicate elements in a mathematical set?  The
short answer is no, but this will be answered further below.  The
important thing is what is IN the set, not how you DESCRIBE the set.
Here is a simple illustration.  Suppose a room contains ten adult
males, ten adult females, ten non-adult males, and ten non-adult
females, for a total of 40 people.  Let's define the set S to be the
set of all people in the room who are either "female" or "non-adult".  
What is the size of that set?  I have carefully worded the definition
to exclude the adult males, so those ten people are disqualified for
membership in the set, leaving 30.
  
If you reason that the ten non-adult females must be counted twice,
then the "head count" in the set S goes like this:

    10  for the 10 adult females, because they are female
    10  for the 10 non-adult females, because they are female
    10  for the 10 non-adult females, because they are non-adult
    10  for the 10 non-adult males, because they are non-adult
   ----
    40  TOTAL

You can't exclude 10 of the 40 and still have 40 left.  That does not
make sense, which is why set membership is defined that way.  The idea
behind this is that there can be many different ways to QUALIFY for
membership in the set, but a particular thing is either in the set, or
not in the set.

Next, is the set [red, blue, green] the same set as [red, green, red,
blue, blue, red]?  The 2 sets of colors you have given both have 3
elements, and they are the same elements, thus they are the same sets.
The word "sequence" is used for when repetition may be important.  If
your examples were considered as sequences, the first would be a
3-element sequence, and the second a 6-element sequence, and they
would be different.

Third, would 1/2 and 2/4 or "black" and "noir" be considered duplicate
elements?  The apparent problem with your black/noir example is
eliminated when you examine what exactly is the nature of the elements
in the set.  Are the elements "colors" or "words"?  If the set is a
set of colors, then the set has one element.  An English speaker would
say that "black is in the set" and a French speaker would say that
"noir est dans l'ensemble".  If the set is a set of words, then the
set has two elements.  "Black" and "noir" are different words, even if
their meaning is the same.
  
For your question about 1/2 and 2/4, we can clarify it in a similar
way.  If you intend these things as rational numbers, then they mean
the same thing.  Once you know that 1/2 is in the set, then you also
know that 2/4 and 1001/2002 and many others are in the set as well.
However, if by 1/2 and 2/4 you mean the character strings "1/2" and
"2/4" which have 3 characters each, that says nothing about the
character string "1001/2002" with 9 characters.

I should comment on your use of the word "allowed".  It is true that
duplicate elements ARE ALLOWED when you specify a set, but duplicates
do not enlarge the set at all.
   
Finally, the intersection of {1/2,1/3} and {2/4,3/9) is the
intersection of 2 identical sets.  For any set A, the intersection of
A and A is A.   
  
I hope this clears it up for you.  It is appropriate to make sure you
know what the definition is, so you can use the concept of set
correctly.  I've also mentioned the related word "sequence".

If you have further questions, please feel free to write back.  

- Doctor Mike, The Math Forum
  http://mathforum.org/dr.math/                                                                                   (Math Forum @ Drexel)




读过了这些确定了是不必重写。也不必重算,但在Probability的时候就不一样,因为次数是重要的。谢谢大家的阅读和指导:)
本帖最后由 grass83 于 24-12-2013 11:06 AM 编辑

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发表于 24-12-2013 11:33 AM | 显示全部楼层
grass83 发表于 24-12-2013 01:21 AM
尚杰,谢谢 可是在Exp2 里,很明显是有5个vowels . 但 却n(F) = 3(因为有重复)。 同样是 n(F) 在prob ...

算probability 和写 set 是两件事。写set是把elements list out(不重复)。算probablity 一定要把全部算进去。
INTERSECTION 可以拿到 vowels = 从12字母抽到任何一个I/E/O。2*E, 2*I, 1*O. 成功率(probablity) 就是5/12

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 楼主| 发表于 24-12-2013 11:53 AM | 显示全部楼层
尚杰 发表于 24-12-2013 11:33 AM
算probability 和写 set 是两件事。写set是把elements list out(不重复)。算probablity 一定要把全部算 ...

好的,谢谢
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发表于 29-12-2013 02:29 AM | 显示全部楼层
要注意的是你如何define你的event set。

从你example 2说起:

(1)
F={vowels from the word "INTERSECTION"}
我们都知道vowels 只有 A,E,I,O,U,所以在"INTERSECTION"的字里的vowel只有E,I,O。
那么n(F)=3

(2)
不过当你算probability的时候,你的event F 就不一样了。
在这里,我们define F={chances to get vowels from the word "INTERSECTION"},
所以n(F)=5


注意(1)和(2)的 set F 的意思是不一样的,这就是之中的差别了。

希望这可以帮得上忙。
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 楼主| 发表于 29-12-2013 10:20 PM | 显示全部楼层
Allmaths 发表于 29-12-2013 02:29 AM
要注意的是你如何define你的event set。

从你example 2说起:

谢谢您的指导:)
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