Why a randomly chosen element of a class satisfying property `p`, implies existence of an element of said class that satisfies `p`?

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I guess I understand that if a randomly chosen element has zero probability of satisfying p (the negation of the conditional), then it does not necessary mean no element exist such that it satisfies p. That explains why a random element of real numbers has zero probability of being an additive identity of R, yet such element exist.

On the other hand I can only justify such implication if one were to equate "zero probability" with "impossibility", which we just said that can't be true (i.e. additive identity of R has zero probability of being selected).

P.S. I'm trying to understand the probabilistic method, and I have not had measure-theoretic probability theory yet.

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First let me just add to your correct consideration that in a finite probability space, where the number of elementary events is finite, if the probability of an event is $0$ then it is not only improbable, it is impossible.

In the probabilistic method the general idea is the opposite of what you are thinking: it consists on defining a probability space where for example the occurrence of a desired property is a specific event. If you then can show somehow that this event has positive probability, then the desired object must exists.

Let me give you a trivial example: suppose we have a bag of colored balls, and I tell you that the probability that you pick a blue ball from it is 0.01. Then you know for certain that there must be at least a blue ball inside the bag.

A classical example is the result from Erdős about lower bound on the Ramsey numbers, see for example https://theoremoftheweek.wordpress.com/2010/05/02/theorem-25-erdoss-lower-bound-for-the-ramsey-numbers/