Probability Problem based on the a set of numbers. Determine if the number is divisible by three Three distinct numbers are selected at random from the set $\{1,2,3,4,5,6\}$. What is the probability that their product is divisible by $3$?
I think that since because $3$ and $6$ are the only numbers that divide into three with an integer remainder, the number would have to have $3$ or $6$ in the unit's digit. Since this is the case the probability would be  $2/6$ or $1/3$. Have I done anything wrong?

We need to find the probability that at least one of the numbers is divisible by three. This is the same as
$$\begin{align*}
1-P(\text{none of the numbers are divisible by three})
&=1-\left(\frac{4}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\right)\\\\
&=1-0.2\\\\
&=0.8
\end{align*}$$
In order to get three numbers that are not $3$ or $6$, we must first select $1,2,4,$ or $5$ out of the six numbers giving a probability of $\frac{4}{6}$. 
Take that number away. We must now select one of the remaining three numbers of the five with probability $\frac{3}{5}$.
Finally, we must now select one of the two remaining numbers of the four with probability $\frac{2}{4}$
