Suppose $X$ and $Y$ are i.i.d. random variables. Also suppose they take the values from the set $\{1,2, \dots, n \}$. Then does this mean that $P(X=1, Y= 1) = P(X=1) \cdot P(Y=1)$ $P(X=1, Y=2) = P(X=1) \cdot P(Y=2) \dots$
So there are are $\binom{n}{2}$ cases for which $P(X \cap Y) =P(X) \cdot P(Y)$? If we didn't know that they were independent, we would have to check all these cases?