Why $P(A|B)\neq P(B|A)$?

Question

In my study of conditional probability, it is given that $$P(A|B)\neq P(B|A),$$ where $A$ and $B$ are two arbitrary events. I was wondering, why is this true? It may seem trivial and counter-examples are easy to produce, but I am hoping for some intuition.

Answer 1

Because $\mathbb P(A\mid B)=\mathbb P(B\mid A)$ is equivalent to $$ \frac{\mathbb P(A\cap B)}{\mathbb P(B)}=\frac{\mathbb P(A\cap B)}{\mathbb P(A)}, $$ which is equivalent to $\mathbb P(A)=\mathbb P(B)$ (assuming that $\mathbb P(A\cap B)>0$), and arbitrary events won't usually satisfy $\mathbb P(A)=\mathbb P(B)$.

Now for the intuitive explanation. Think about what $\mathbb P(A\mid B)$ means, in terms of a Venn diagram: you are shrinking the probability space down to just the set $B$, and looking to see what proportion of $A$ still falls inside that, but it is relative to the whole size of $B$. If it wasn't for the "relative" part, then it wouldn't matter if you started with $B$ and shrunk $A$ down, or started with $A$ and shrunk $B$ down. But since conditional probabilities are relative, it matters if the space you are "shrinking down to" is $A$ or if it is $B$, since that determines what quantity you are dividing by in the conditional probability.