Axiom of Probability: Question on p(<sample space>)=1

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I have a question regarding one of the axiom for probability, which is p(<sample space>)=1.

I do not understand why p(<sample space>)=1 is an axiom instead of theorem, since I can prove it with the following argument:

Since sample space has been defined as the set of all possible outcomes, hence in any case, the sample set must occur, therefore the probability of the occurring of sample space is 100%, which follows that p(<sample space>)=1.

What is the problem with the argument?
 
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Is the fact that something that must occur has probability 100% (i.e. 1.0) a separate axiom? If so, how do they define "must occur"?
The point is that the idea that something that "must occur" has probability 1.0, is derived from the axiom you cite, not vice-versa.
 
Thanks, but why we do not called
p(<sample space>)=1
as a definition instead of an axiom?

I am not sure what make us call it an axiom...
 
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What would it be defining? It is stating a property, not defining a word. If you are thinking of that as defining "must happen", that phrase is not part of mathematical probability but rather of possible applications of probability.

(This may beyond what you are studying, but if your sample space is "all numbers between 0 and 1" and all numbers are equally likely, then the probability you will select an irrational number is 1, even though it is possible to select a rational number.)
 
Thanks again :)
The reference book I have used stating that:
Axiom 1 stating that 0<=P(E)<=1
Axiom 2 stating that P(S)=1
Axiom 3, the probability of union of mutually exclusive events is equal to the summation probability of of each of the events.

And the author says that, hopefully, the reader will agree that the axioms are natural and in accordance with our intuitive concept of probability as related to chance and randomness.

But what if axiom 1 and axiom 2 is changed to
Axiom 1 stating that 0.5<=P(E)<=1.5
Axiom 2 stating that P(S)=1.5
(Axiom 3 no change)

or

Axiom 1 stating that 1.1<=P(E)<=2
Axiom 2 stating that P(S)=2
(Axiom 3 no change)

and rebuild the probability model base on the new axiom? Will there be any problem in this new probability model?

If not can I say that the original Axiom 1 and Axiom 2 is just taking some reference value so everybody on the Earth can follow it?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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