I would like to know why the 'solution' with which I came up is wrong. I gave the correct answer at the bottom.
Each ball has probability $\frac{1}{3}$ of going into either bin, and there are a total of $3^3=27$ ways to distribute the balls among the bins.
My solution: There are $3$ ways to select the bin for the first ball, $2$ ways to select the bin for the second ball (because it must not go into the bin selected for the first ball), and $2$ ways to select the bin for the third ball (because it must go into either of those selected before). This gives a probability of $\frac{3\cdot 2\cdot 2}{27}=\frac{12}{27}=\frac{4}{9}$.
Correct answer: $\frac{{{3}\choose{2}}(2^3-2)}{27}=\frac{2}{3}$, because there are ${3}\choose{2}$ ways to select the bins into which the balls should go, $2^3$ ways to put the $3$ balls into these bins, and in $2$ of those cases all balls are in the same bin.
As I wrote in the opening, I don't quite see what's wrong with the first attempt.
Best Answer
You exclude one possibility in your solution. The possibility you are excluding is;
If the second ball goes into the same bin as the first ball, and the third ball goes into another one. The calculation is as follows:
There are $3$ ways to select the bin for the first ball, $1$ way to select the bin for the second(as it will be the same bin as the first ball), and $2$ ways to select the bin for the third ball. This gives the probability of $\dfrac{3\cdot{1}\cdot{2}}{27} = \dfrac{2}{9}$ If you add this probability to the probability you computed, you will also get $\dfrac{2}{3}$