Why “probability of 0” does not mean “impossible” | Probabilities of probabilities, part 2

What people said

Recommends user765539 · 2020-04 · Mathematics Stack Exchange · on “What does measure zero mean in probability?”

$\def\Fcal{\mathcal{F}}$ But setting aside the "real-world" argument, if we're talking about probability theory in the context of measure theory, what does $P(A) = 0$ mean? In the context of measure theory, a fundamental concept in probability theory is called probability space [LINK] , which consists of three elements: A sample space $\Omega$ ; A $\sigma$ -field $\Fcal$ on the set $\Omega$ ; A probability measure $P:\Fcal\to[0,1]$ . For any element $A\in\Fcal$ , which is often called an event , it is associated with a unique real number $P(A)$ in $[0,1]$ . The identity $P(A)=0$ simply means "the probability measure of the event $A$ is zero". If $A$ isn't impossible, what does $0$ mean, if a

Recommends Mark S. · 2020-04 · Mathematics Stack Exchange · on “How can an event with probability $0$ be possible?”

Rather appropriately, right around the time of my previous comment popular math Youtuber Grant Sanderson [LINK] , aka 3Blue1Brown [LINK] just posted a Youtube video titled "Why 'probability of 0' does not mean 'impossible'" [this video] , which I bet might be helpful, at least a bit.

Recommends awkward · 2020-07 · Mathematics Stack Exchange · on “Dividing by infinities while doing ratio of areas in probability”

You might be interested in this video by 3blue1brown in which he discusses probability density functions: [this video]

Recommends Sigma · 2020-09 · Mathematics Stack Exchange · on “Probability calculation with an infinite variable?”

I believe that the event of choosing the correct number is a possible event, but it's probability is still 0. Watch this 3B1B video: [this video]

Cites user151496 · 2014-05 · Mathematics Stack Exchange · on “Choosing something of $0$ probability”

First of all, i am only a newbie and i am pretty sure that my thinking is faulty somewhere, so along the way as i explain things i will probably say something with an error and i hope you will help me find this error. This idea has been bogging me for some time already. Imagine picking a random real number. Without the loss of generality, let the picking interval be $(0,1)$ , or $<0,1>$ . Now, since there is an infinite amount of real numbers, the probability of choosing one specific number is $0$ . Yet, we have to pick one number when choosing randomly, let's say it's $0.5$ . We have now picked a number that had $0$ chance to be picked. How is this possible? Thank you for all the answers, i

Cites masiewpao · 2020-04 · Mathematics Stack Exchange · on “Why we can draw a random number $s$ in a normal distribution when we are taught that $p(s)=0$?”

I think the question has two main aspects: The first is the practical implementation. As pointed out in the comments, no programme will give you a decimal of infinite precision; you have to be content with some fixed precision floating point number. These numbers are not truly random, they are instead pseudo-random. They use a particular algorithm to generate successive floating point numbers. In python I believe the primary implementation is the Mersenne Twister algorithm (per the docs). The important things to realise here is that pseudo-random number generators are completely deterministic; you just need the seed value, and that they have fixed precision. The other aspect to this question

Cites user35885 · 2020-04 · Mathematics Stack Exchange · on “Is the probability of zero is not impossible according to this YouTube video link?”

I found the following link on youtube.com: Probability of zero is not impossible [this video] $\text{Isn't } \textbf{P(0)} = 0 \;\;?$ I did not understand this video. Can anyone kindly tell me why that is so? Why probability of zero is not impossible? I am asking the above question because in the following link at the bottom of the website, one of the poster said the probability of zero means impossibility. Then? Where is my error in understanding? I am really confused now. Can anyone kindly elaborate on this paradoxical math problem or is it really a paradox? Probability of zero means impossible [LINK]

Cites K.defaoite · 2020-09 · Mathematics Stack Exchange · on “What is the formal definition of "probability distribution"?”

1: Formal definitions To start with this question, one should define a probability space : A tuple of three items usually denoted $(\Omega,\mathcal{E},\Bbb{P})$ [or something of this nature]. $\Omega$ is the sample space - the set of all possible outcomes (not to be confused with events !) of our procedure, experiment, whatever. For instance, consider flipping a coin once: In this case, $\Omega=\{\text{H},\text{T}\}$ . A random variable $X$ is the "result" of this experiment. You could define $X$ in this case as $$X=\begin{cases} 1 & \text{If coin lands heads}\\ 0 & \text{If coin lands tails} \end{cases}$$ Formally, one can define a measurement $M$ as a bijective map $M:\Omega\to\mathcal{X}$

Cites PrincessEev · 2024-01 · Mathematics Stack Exchange · on “How Can Infinite Sums of Zero Sum to 1?”

So again, infinite sums of 0 appear to be equal to a non-zero result. Not quite. Your focus was on $\frac{b-a}{n}$ , but $n$ appears two times in that expression: the second time is in the upper bound of the summation: $$ \frac{b-a}{\color{blue}{n}} \sum_{i=1}^{\color{blue}{n}} f(x_i) $$ So while the factor out front shrinks, you're summing up even more values of the function. For well-behaved functions, these two phenomena "destructively interfere" in a sense, usually reaching a finite value and often nonzero. For others, it can blow up altogether. (To see why, note that the integral can often be thought of as the area under a curve.) In general, the logical flaw here is the claim $$ \lim_{