I came across a very interesting probability question which I intuitively know the answer to, but I cannot come up with an example that satisfies my answer.
If $P(A)+P(B)>1$, find the minimum value of $P(A \cap B)$.
My hunch is that the minimum value is $0$, as although in the probability space events $A$ and $B$ have to intersect, the question doesn't specify how big the intersection should be, and so the intersection can be made arbitrarily small as possible, and thus $P(A \cap B) \rightarrow 0$ as you make the intersection smaller and smaller.
But does really making the intersection arbitrarily small make $P(A \cap B)=0?$ I considered the uniform distribution (a classic case would be throwing darts at the uncountably infinite real line).
For the uniform distribution, $P(X=k)=0$, where $k$ is some value. Consequently, any countably finite set of values $P(X=\{k_1, k_2,...,k_n\})=0$. If you want a non-zero probability, you have to take an interval instead of a discrete set of values.
For my above example, I considered $P(X\leq0.5+\epsilon)$ and $P(X>0.5)$, where $\epsilon$ is some very small value. The sum of both probabilities is slightly greater than $1$, but can I really say that $P(X\leq0.5+\epsilon\ \cap X>0.5)=P(0.5<X\leq 0.5+\epsilon)=0?$ After all, no matter how small $\epsilon$ is, the event $0.5<X\leq 0.5+\epsilon$ is still an interval, and it should have non-zero probability (at least for the uniform distribution).
Can someone help resolve my confusion?
Your example with the uniform distribution shows that $P(A \cap B)$ can be arbitrarily close to $0$.
To see that it cannot equal zero, note that if $P(A \cap B) = 0$ then $P(A) + P(B) = P(A \cup B) - P(A \cap B) = P(A \cup B) \le 1$ by inclusion-exclusion.
So $P(A \cap B)$ can be as close to $0$ as you like, but not equal $0$.