That's the from stat110 book, problem 23
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?
I have found several threads with different solutions here, but none with mine.
My solution:
Strategy: using the complement, $1-$ what is the probability that all robberies happen in different district?
The first robbery can happen in any district out of the six.
The next robbery can happen in any of the $5$ other districts, then there are $4$, $3$, $2$, $1$ districts.
which translated to me into a probability of $$1 – \frac{5}{6}\frac{4}{6}\frac{3}{6}\frac{2}{6}\frac{1}{6}$$ or $$1 – \frac{5!}{6^5}$$
the solution is $$1 – \frac{6!}{6^6}$$
Where did my reasoning fail?
thank you very much for explaining!
Best Answer
$$\frac{5!}{6^5} = \frac{6}{6} \cdot \frac{5!}{6^5} = \frac{6!}{6^6}.$$