What is the probability of three union if one of the events has probability zero?

Question

my problem is this: We throw $n$ bells randomly into $3$ boxes initially empty. Compute the probability that at least One boxes remaining empty. I have the solution and that is: $$ P(A_1 \cup A_2 \cup A_3) = P(A_1) + P(A_2) + P(A_3) - P(A_1 \cap A_2) - P(A_3 \cap A_2) - P(A_1 \cap A_3) + P(A_1 \cap A_2 \cap A_3) $$ $$ P(A_1)=(2/3)^n $$ $$ P(A_1 \cap A_2) = 1/3^n $$ $$ P(A_1 \cap A_2 \cap A_3) = 0 $$ But question is: why the probability of the intersection of two is not $0$? In other words, why the probability of the intersection of three is $0$ but the intersection of $2$ is not $0$? Aren't they $3$ disjoint events? How is it possible that a ball is in $A_1$ AND $A_2$?