Five standard 6-sided dice are rolled. what is the probability that at least 3 of them show a $6$?
I have $(\frac{1}{6})^3(\frac{5}{6})^2+(\frac{1}{6})^4(\frac{5}{6})+(\frac{1}{6})^5=\frac{31}{7776}$, but this is wrong, and the actual solution uses PIE (principle of inclusion and exclusion) and seems like it's more complicated that it needs to be. This is a very simple problem, so I'm not sure what I did wrong. I don't believe I need to use PIE, because I can just multiply probabilities like I did above. What did I do wrong?
$$P(\text{at least 3 of them show 6}) = \sum_{5 \geq i \geq3} P(\text{ i of them show 6})$$ $$P(\text{i of them show 6}) = {5 \choose i } \left(\frac{1}{6}\right)^{i}\left( \frac{5}{6}\right)^{5-i} $$
EDIT
I understand what you asked now, so you can see the error you are making is by not adding the combinatorial coefficients, which means you are not treating the distinguishability of the die's. For example in your treatment, you are treating $P(\text{3 rolls of 6})$ as the probability of an event x.x.x.6.6.6, but you are not adding, x.6.x.x.6.6 and other possible combinations, Is it clearer now ?