Zero probability and impossibility I read a comment under this question:


  There are plenty of events that can occur that have zero probability.


This reminds me that I have seen similar saying before elsewhere, and have never been able to make sense out of it. So I was wondering 


if zero probability and impossibility mean the same?
if an event with zero probability doesn't mean that the event is impossible to occur, how probability theory represents/describes impossibility?


Thanks and regards!
Two Schools

I think the crux of the matter is what probability actually is:


The Bayesian view - probabilities are measures of (personal) confidence or belief, so it's quite obvious why an event with probability zero is not the same thing as an impossible event. But perhaps this isn't such a satisfactory answer.
The frequentist view - probabilities are the asymptotic frequency of events as the number of independent trials tends to infinity. Here again wee see that something that happens with probability zero is not the same as something impossible; it's just something that happens so infrequently that the numerator in $\dfrac{\text{occurences}}{\text{trials}}$ is dominated by the denominator.


Technically Speaking

Putting aside such philosophical matters, there's also a technical matter to be discussed here. Under the usual measure-theoretic formulation of probability theory, we have a sample space $\Omega$ and a family $\mathcal{F} \subseteq \mathcal{P}(\Omega)$ of events (measurable subsets of $\Omega$), and the probability of an event $A \in \mathcal{F}$ is its measure $\mathbb{P}(A)$. There is nothing in the axioms of measure theory which say that a non-empty set must have a non-zero measure; and if we interpret $\mathcal{F}$ as the set of all possible events, it's clear that an impossible event is not the same thing as an event of zero probability.

Example

To give a concrete example, consider a random variable $X$ which is uniformly distributed on the interval $[0, 1]$. Although $\mathbb{P}[X \in (a, b)] = b - a$ for all $(a, b) \subset [0, 1]$, the axioms of probability force us to conclude that $\mathbb{P}[X = x] = 0$ for any individual $x \in [0, 1]$: for if $\mathbb{P}[X = x] = \varepsilon &gt; 0$, because $X$ is uniformly distributed, by additivity of the probabilities of disjoint events, we'd be forced to conclude that $[0, 1]$ contains at most $\frac{1}{\varepsilon}$ (a finite number!) points, which is absurd.
I don't know if this is correct, but this is how I made sense of it, I think this a very intuitive explanation. 

Imagine a car moving (from left to right) in a straight line with constant speed of 100MPH, and at x=0 the breaks of the car are engaged. Naturally the car will start to decelerate until it comes to rest (speed = 0 MPH).

The car will certainly come to rest, but the question is:

At what distance D [meters] (with respect to x=0) will the car come to rest? (This value D is not known apriori and can vary from one experiment to the next by a myriad of factors, so we can ask probability questions about the variable D).

Since distance is modeled by real numbers, the variable D is continuous. If we ask:
what is P(D=6.354), the probability that it comes to rest at specifically D = 6.354 meters?

then, since the probability distribution of D is continuous, the answer is P(D=6.354)=0. This probability is zero.

Notice that there is nothing special about the number "6.354". That is, for any real number r the probability that D is exactly r is zero, i.e. P(D=r) = 0. This is simply a consequence of how continuous probability distributions are defined.

Regardless, we know that the car will certainly come to rest and that this happens at some specific distance d (where d is a real number). So even though P(D=d) = 0, it does not mean that coming to rest at D=d was impossible; clearly it was not because the car does come to rest at d.