If the probability of A = 0, it doesn't mean A is an empty set I am new to stats and I have a question regarding the following statement: If Pr(A) = 0, it doesn’t mean that event A is an empty set.
However, isn’t the probability of A = (number of sample points in A) / (number of sample points in the same space) and that events are subset of the sample space.
Thus, based on the formula above, the only way the Probability of A is 0 is if the number of sample points in A is 0 thus making it an empty set. Furthermore, A cannot contain elements that are outside the sample space thus A would be an empty set.
However, this is contradictory to the statement If Pr(A) = 0, it doesn’t mean that event A is an empty set.
 A: The rule that $\mathbb{P}(A) = |A|/|\Omega|$ (for sets $A\subseteq \Omega$) arises in a specific probability context, where you have a sample space $\Omega$ containing a finite number of outcomes all with equal probability.  In that particular context, you you are correct that $\mathbb{P}(A)=0$ only when the set $A$ is empty.  However, probability theory deals with much more general situations than this, so that rule holds only in a very narrow class of cases.
More generally, probability theory deals with cases where there are events $A$ that are non-empty sets of outcomes, but they still have zero probability.  This leads us to make a distinction between events that are certain to occur and events that are merely almost sure (i.e., occur with probability one).
A: $A$ doesn't have to be an empty set. For instance, consider all living people queuing in a straight line. So $S = \{ \text{ all living people } \}$. Let $B= \text{ You are at the front of the line } $.
Now $P(B)\approx 1/7\text{billion} = 1.4 \times 10^{-10}$. Now consider the set $T = \{ \text{ all dead people } \}$. Note that $S \cap T = \emptyset$ (the intersection of $S$ and $T$ is empty). The probability a dead person (say, $A = \text{Genghis Khan}$) is as the front of the line is $0$. This is because $A \cap S = \emptyset$.
In short, $P(A) = 0$ doesn't imply $A = \emptyset$ but rather $A \cap S=\emptyset$.
