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Independance

Poll

Do you wish for Probability and Statistics to be independant of the Mathematics forum

I would like to propose the following:

Creating a 'Probability and Statistics' sub-forum separate from the 'Mathematics' sub-forum.

I speak for the many of us who would prefer to have their own building (or sub-forum in this case) so we can avoid other mathematicians if we wish.:smile::biggrin:
Reply 1
Avatar for Notnek
Notnek
21
Original post by tombayes
I would like to propose the following:

Creating a 'Probability and Statistics' sub-forum separate from the 'Mathematics' sub-forum.

I speak for the many of us who would prefer to have their own building (or sub-forum in this case) so we can avoid other mathematicians if we wish.:smile::biggrin:

Not enough demand - read through all the thread titles in the maths forum.

Also, some of the other subforums have posts which go unanswered for hours/days. But here, most maths questions get answered pretty quickly including stats questions. A new subforum wouldn't have enough helpers.
Reply 2
Avatar for Zacken
Zacken
22
I don't think that would be a particularly good idea. The subset of statistics and probability questions in the set of math questions that get asked here is too small to require another sub forum.

Plus, I saw we Mathematicians unite. :biggrin::wink:
Reply 3
Avatar for tombayes
tombayes
OP
12
Original post by Zacken

Plus, I saw we Mathematicians unite. :biggrin::wink:


of course we would occasionally visit the maths forum when you guys are struggling:biggrin:.
Original post by notnek
Not enough demand - read through all the thread titles in the maths forum.


This is why I voted no


Though to be frank I despise Stats so much that I can see a real plus
Reply 5
Avatar for TeeEm
TeeEm
19
Original post by tombayes
I would like to propose the following:

Creating a 'Probability and Statistics' sub-forum separate from the 'Mathematics' sub-forum.

I speak for the many of us who would prefer to have their own building (or sub-forum in this case) so we can avoid other mathematicians if we wish.:smile::biggrin:



As much as I do not particularly like statistics and probability (and possibly pure mathematics) I still enjoy looking, though I may not contribute, to these threads.

Is this like Scotland independence from the UK?
Reply 6
Avatar for tombayes
tombayes
OP
12
Original post by TeeEm
As much as I do not particularly like statistics and probability (and possibly pure mathematics) I still enjoy looking, though I may not contribute, to these threads.

Is this like Scotland independence from the UK?


unlike Scottish independence this will actually happen:biggrin:
Reply 7
Avatar for TeeEm
TeeEm
19
Original post by tombayes
unlike Scottish independence this will actually happen:biggrin:


I will personally miss your contributions ...:tongue:
Original post by tombayes
I would like to propose the following:

Creating a 'Probability and Statistics' sub-forum separate from the 'Mathematics' sub-forum.

I speak for the many of us who would prefer to have their own building (or sub-forum in this case) so we can avoid other mathematicians if we wish.:smile::biggrin:


Why should there be any subforums for maths other than the dedicated exams one?

All that introducing more subforums would do is fragment the maths section and make it more likely that threads will go unnoticed by somebody who could have helped.
Let A be the event that someone studies mathematics.
Let B be the event that they study probability / statistics.

We know that B⊂AB \subset A. We also know that not everyone studies mathematics, so P(A)<1\mathbb{P}(A) < 1.

Now note that since A∩B=BA \cap B = B, we have:

P(A∩B)=P(B)>P(A)P(B)\mathbb{P}(A \cap B) = \mathbb{P}(B) > \mathbb{P}(A)\mathbb{P}(B) (since P(A)<1\mathbb{P}(A) < 1).

That is, P(A∩B)≠P(A)P(B)\mathbb{P}(A \cap B) \neq \mathbb{P}(A)\mathbb{P}(B), and so A and B are not independent.

We deduce that the only correct answer to the poll question is "No, they should not be independent". You can't argue with the maths!
(edited 9 years ago)
Original post by DFranklin
...


Nice one. PRSOM.
Reply 11
Avatar for tombayes
tombayes
OP
12
Original post by DFranklin
Let A be the event that someone studies mathematics.
Let B be the event that they study probability / statistics.

We know that B⊂AB \subset A. We also know that not everyone studies mathematics, so P(A)<1\mathbb{P}(A) < 1.

Now note that since A∩B=BA \cap B = B, we have:

P(A∩B)=P(B)>P(A)P(B)\mathbb{P}(A \cap B) = \mathbb{P}(B) > \mathbb{P}(A)\mathbb{P}(B) (since P(A)<1\mathbb{P}(A) < 1).

That is, P(A∩B)≠P(A)P(B)\mathbb{P}(A \cap B) \neq \mathbb{P}(A)\mathbb{P}(B), and so A and B are not independent.

We deduce that the only correct answer to the poll question is "No, they should not be independent". You can't argue with the maths!


:hide::hide: hahaha.

However, I would disagree:

We know that B⊂AB \subset A. Really? - What about biologists, and social science people who do have modules in stats but not maths.

However, if you assume the above to be true then:

As before:

Let A be the event that someone studies mathematics.
Let B be the event that they study probability / statistics.
And let MM be the event someone asks or answers a question on the maths forum. And lets us only consider people who are on the maths forum so P(M)=1P(M)=1 and P(A)=1P(A)=1. (Assuming everyone on the maths forum studies maths).

Now note that since A∩B=BA \cap B = B, we have:

P(A∩B)=P(B)=P(A)P(B)P(A \cap B) = P(B) = P(A)P(B) (since P(A)=1P(A) = 1).
so A and B are are independent.
Original post by tombayes
:hide::hide: hahaha.

However, I would disagree:

We know that B⊂AB \subset A. Really? - What about biologists, and social science people who do have modules in stats but not maths.

However, if you assume the above to be true then:

As before:

Let A be the event that someone studies mathematics.
Let B be the event that they study probability / statistics.
And let MM be the event someone asks or answers a question on the maths forum. And lets us only consider people who are on the maths forum so P(M)=1P(M)=1 and P(A)=1P(A)=1. (Assuming everyone on the maths forum studies maths).

Now note that since A∩B=BA \cap B = B, we have:

P(A∩B)=P(B)=P(A)P(B)P(A \cap B) = P(B) = P(A)P(B) (since P(A)=1P(A) = 1).
so A and B are are independent.


Since there are a number of Maths teachers/tutors etc on here, not everyone studies Maths, i.e. P(A) < 1
Reply 13
Avatar for tombayes
tombayes
OP
12
Original post by tiny hobbit
Since there are a number of Maths teachers/tutors etc on here, not everyone studies Maths, i.e. P(A) < 1


i consider doing maths and teaching maths the same as studying maths.

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