Why is the probability of $\emptyset$ equal to 0? I'm wondering why is the probability of the empty set, $\emptyset$, equal to 0? 
\begin{equation}
P(\emptyset) = 0.
\end{equation}
Isn't the empty set always included when you take the all the possible subsets of a set?
 A: Suppose you pick a number between $1$ and $10$ at random. What's the probability that the number is even? $1/2$. What's the probability that you picked $1$ or $2$? $1/5$. Now, what's the probability that the number you picked is neither even nor odd?
A: I think a more rigorous proof than previously given as answers uses more axioms of a probability measure. Of the three axioms, we need:

*

*$\mathrm{P}(\Omega)=1$.


*Given $A_i$ are mutually disjoint events in a sample space $\Omega$, $\mathrm{P}\left(\cup_{i=1}^{\infty}A_i\right)= \sum_{i=1}^{\infty}\mathrm{P}(A_i)$
So we say $A_i = \emptyset$, because these are mutually disjoint, we can use axiom 2, such that:
$$\mathrm{P}\left(\cup_{i=1}^\infty A_i\right)=\sum_{i=1}^{\infty}\mathrm{P}(A_i)=\mathrm{P}(\emptyset)+P(\emptyset)+\dotsb.$$
However, we know from axiom one that the probability of a event must be finite from axiom 1. Therefore $\mathrm{P}(\cup_{i=1}^\infty A_i)<\infty$ is true only if $\mathrm{P}\left(\emptyset\right)=0$.
A: It's a consequence of the axioms of a probability measure: if $A$ is any event, then $A\cup\emptyset=A$ and $A\cap\emptyset=\emptyset$, hence
$$ \mathbb{P}(A)=\mathbb{P}(A\cup\emptyset)=\mathbb{P}(A)+\mathbb{P}(\emptyset)$$
Therefore we must have $\mathbb{P}(\emptyset)=0$.
