Zero probability vs impossibility

[AvgStudent]

My professor told me today the probability of 0 does not mean the event is impossible. I’m trying to wrap my head around this. If probability of 0 does not imply impossibility, then what is the probability of 0? If the probability of 0 is different than impossibility, then how do you represent impossibility?

 

[blamocur]

Consider any continuous distribution, e.g., normal. Pick a single value from the distribution. The probability of picking this value is 0.

 

[Cubist]

Consider any continuous distribution, e.g., normal. Pick a single value from the distribution. The probability of picking this value is 0.

A very interesting question and answer. I wasn't aware of this. I just did some internet searches on the topic. This has obviously been contemplated by some very clever people. It reminds me of the controversy that surrounded calculus until the introduction of limits, ie there seem to be "devils in the details".

 

[Deleted member 4993]

My professor told me today the probability of 0 does not mean the event is impossible. I’m trying to wrap my head around this. If probability of 0 does not imply impossibility, then what is the probability of 0? If the probability of 0 is different than impossibility, then how do you represent impossibility?

Suppose you have 7 balls lined up in a straight. You want to pick the middle ball - with eyes wide-open. The probability of success is almost ~ 1.

Suppose you drew a line 2" long. Now, you want to mark the mid-point, just with a pencil. The probability of success is 0 (improbable), but it is not impossible.

 

[pka]

A very interesting question and answer. I wasn't aware of this. I just did some internet searches on the topic. This has obviously been contemplated by some very clever people. It reminds me of the controversy that surrounded calculus until the introduction of limits, ie there seem to be "devils in the details".

Surely this idea was discussed in any university level course in probability??
Consider [imath]X[/imath] being uniformly distributed on [imath](-1,1)[/imath]
Then [imath]{f(x)=\begin{cases}f(X)=\dfrac{1}{2} & X\in (-1 ,1)\\ 0 &\text{ otherwise}\end{cases}}[/imath]
Now it is possible that [imath]X=0[/imath] but [imath]\mathcal{P}(X=0)=0[/imath]

 

[Cubist]

Surely this idea was discussed in any university level course in probability??
Consider [imath]X[/imath] being uniformly distributed on [imath](-1,1)[/imath]
Then [imath]{f(x)=\begin{cases}f(X)=\dfrac{1}{2} & X\in (-1 ,1)\\ 0 &\text{ otherwise}\end{cases}}[/imath]
Now it is possible that [imath]X=0[/imath] but [imath]\mathcal{P}(X=0)=0[/imath]

I don't remember it being discussed and I went to a very good university. It's certainly not impossible that I was hung over on that day . Thanks for your explanations!