# Is it possible for an event $A$ to be independent from event $B$, but not the other way around?

I was wondering if event $$A$$ is independent of event $$B$$, would $$B$$ also be independent of event $$A$$? My original thought was that it should be independent, but then I realized if $$A$$ is independent of $$B$$ then we have:

$$P(A|B)=P(A)\label{1}\tag{1}$$ and for $$B$$ to be independent from $$A$$ we need to have: $$P(B|A)=P(B)\label{2}\tag{2}$$

but in $$\ref{1}$$ if $$P(A)=0$$ then $$\ref{2}$$ doesn't make sense, so then $$B$$ wouldn't be independent of $$A$$?

Thank you

$$P(B\mid A)$$ is undefined when $$P(A)=0$$, so you can’t draw any conclusions about independence of the two events from it. That one reason why (despite what the Wikipedia page on conditional probability might imply) the fundamental definition of independence of two events uses their joint probability: $$A$$ and $$B$$ are independent iff $$P(A\cap B)=P(A)P(B)$$. This definition is symmetric.
$$P(A \mid B) = P(A)$$ should not be taken as the definition of independence, $$P(A \cap B) = P(A)P(B)$$ should be taken as the definition of independence. From this we can prove $$P(A \mid B) = P(A)$$ as a corollary, provided that $$P(B) > 0$$.
$$P(A|B)=P(A)$$ is not the correct definition of independence. The correct definition is $$P(A\cap B)=P(A)P(B)$$. These definitions are equivalent if $$P(B)>0$$. With the correct definition there is symmetry between $$A$$ and $$B$$ so $$A$$ independent of $$B$$ is same as $$B$$ independent of $$A$$