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In many problems of probability the requirement is : Find the probability of (something). But a probability is a positive numeric function s.t:

Axiom 1. $P(E)=1$ ,where $E$ is sample space.

Axiom 2. Let $A_1,A_2,\ldots$ be a countable (possibly countably infinite) sequence of pairwise disjoint events. then: $$ P\left(\bigcup_n A_n\right)= \sum_n P(A_n)$$

But there are many functions, satisfying these axioms, so what does it mean to find the probability of some event? Or is the probability somehow unique?

gt6989b
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Ica Sandu
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    This question usually means, given a probability function $P$ and a specific event $E$, determine $P(E)$. The probability function is already defined, explicitly or implicitly. – eepperly16 May 02 '19 at 18:01
  • Of course you need to work in a specific probability space. – Mark May 02 '19 at 18:04
  • Do you have a specific problem in mind? I expect you'll see that the probability distribution is generally stated explicitly. – lulu May 02 '19 at 18:13
  • But then I can define a probability which not return how frequent an event occurs.then for what is good my probability? – Ica Sandu May 02 '19 at 18:16

3 Answers3

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Good question - no the probability is not unique. It is generally a measure defined on the sample space $\Omega$, which must be equipped with a sigma algebra $\mathcal{F}$. The pair $(\Omega, \mathcal{F})$ is called a measurable space (it the sense that it admits a measure - here, a probability measure).

However, when we ask what is the probability that $X \in A$ (e.g., $P[X\leq 1]$), what we actually measure is the subset of $\Omega$, $X^{-1}(A)$.

Rewind: Let's take it form the beginning: $(\Omega, \mathcal{F})$ is a measurable space. We equip it with some probability measure $P:\mathcal{F} \to [0,1]$ (which satisfies the axioms you mentioned). Let $(E, \mathcal{E})$ be another measurable space - take for example $\mathbb{R}$ with its Borel subsets.

A function $X:\Omega \to E$ is called a random variable if it is measurable, that is, if $X^{-1}(C) \in \mathcal{F}$ whenever $C\in \mathcal{E}$.

The "probability that $X$ is in $A$", denoted by $P[X\in A]$, is a shorthand for

$$ P[X^{-1}(A)] = P[\{\omega \in \Omega : X(\omega) \in A\}]. $$

This explains why we required that $X$ should be measurable: if $X$ is measurable, $\{\omega \in \Omega : X(\omega) \in A\}$ is measurable whenever $A$ is measurable.

The value of this probability depends on the choice of $P$ on $(\Omega, \mathcal{F})$. This is certainly something one needs to clarify. Sometimes probability measures are defined in terms of probability distributions, that is, functions $F_X(x) = P[X\leq x]$. This definition does not allow us to measure the sets of $\mathcal{F}$ directly (but we usually don't care), but permits to probe into $\Omega$ using $X$.

I hope this answers your question to some extent.

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I think you are generally confused about particular usage of the word probability.

There are multiple possible probability distribution functions which satisfy the axioms you quoted. For example, both standard uniform and standard normal distributions are defined using such functions, and they assign different weights to different parts of the probability space.

In this context, you can look at multiple distribution functions, if $X \in \{1,2\}$ and $X=1$ with probability $0.25$, what is the probability of $X=2$? (You are essentially trying to compute the actual distribution function for the random variable $X$.)

However, in the context of the question to find the probability of some event, a specific probability function is usually implied by the question itself. For example, find the probability that $-1 < X < 1$, if $X$ is a standard normal random variable.

gt6989b
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In general you have some "axiomatic" views about certain probabilities, e.g., that the probability of a ${\tt head}$ on a coin toss is ${1\over2}$, or that the probability of a ${\tt 1}$ in a die toss is ${1\over6}$. Given these axioms probability theory then tells you how to compute the probability of more complicated events, e.g. to obtain three consecutive ${\tt head}$s in $10$ coin tosses.