I was trying to solve this problem (taken from Harvard's Stat 110 class) using the inclusion-exclusion principle:
A city with 6 districts has 6 robberies in a particular week. Assume the robberies are located randomly, with all possibilities for which robbery occurred where equally likely. What is the probability that some district had more than 1 robbery?
Here's how I tried to approach the problem:
Define $A_i$ as the event that more than $1$ robbery occurs in district $i$, with $ 1 \leq i \leq 6$.
Then, what we need to find is $P(A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6)$.
By the inclusion-exclusion principle, this is equal to $\Sigma_{i=1}^{6}P(A_i)$ $-$ $\Sigma_{i<j} P(A_i \cap A_j)$ $+$ $\Sigma_{i<j<k} P(A_i \cap A_j \cap A_k)$ (the terms involving more than $3$ $A_i$s are not possible because there are only 6 robberies, and we can't have more than $1$ of them, when there are $>3$ districts).
Now, $\Sigma_{i=1}^{6}P(A_i) =$ $6 \times \frac{5}{6}$, since $2,3,4,5,$ or $6$ robberies can happen in each of the $6$ districts.
Also, $\Sigma_{i<j} P(A_i \cap A_j)$ = $6\choose2$ $\times$ $\frac{6}{36}$, since for each of the $6\choose2$ pairs of districts, the robbery counts can be $\{(2,2),(2,3),(2,4),(3,2),(3,3),(4,2)\}.$
Counting in a similar fashion, $\Sigma_{i<j<k} P(A_i \cap A_j \cap A_k) = $ $6\choose3$ $\times$ $\frac{1}{216}$.
But adding the above three terms gives me a probability greater than $1$. Where am I going wrong?