Probability of two die not rolling $1$ or $2$. Probability of two dice not rolling $1$ or $2$.
Let $A$ be the probability of rolling $1$
Let $B$ be the probability of rolling $2$
My thoughts. It is the same as 
$$1-P(A \text{ or } B)$$
$$P(A \text{ or } B) = P(A)+P(B)-P(A \text{ and } B)$$
where
$$P(A)=P(B)=1/6$$
$$P(A \text{ and } B)=1/36$$
Is this right thinking?
 A: @lulu points out a very easy way to solve this problem. However, your way of solving it is more interesting, so let's keep going with this logic.
What you did wrong is that you set the events of $A$ and $B$ wrong. $A$ should have been the event in which the first die rolled a $1$ or a $2$ and $B$ should have been the event in which the second die rolled a $1$ or a $2$. Thus, we have the following:
$$P(A)=P(B)=\frac 2 6=\frac 1 3$$
$$P(A \ \text{and} \ B)=P(A)*P(B)=\frac 1 9$$
$$1-P(A \ \text{or} \ B)=1-(P(A)+P(B)-P(A \ \text{and} \ B))=1-\left(\frac 1 3+\frac 1 3-\frac 1 9\right)=1-\frac 5 9=\frac 4 9$$
The last line gives us our answer.
A: Your reasoning is logically okay, but the numbers you are using are wrong. The odds of rolling a $1$ with two dice is not $1/6$, rather it is $11/36$. Instead of having the events be the numbers, they should be the rolls. So, $A$ is the first die being a $1$ or a $2$ and $B$ is the second die being a $1$ or a $2$. These both have $p=1/3$. Then applying your equations will yield the right answer.
This is one of their cases where counting comes greatly in handy.
There are six ways to roll $(1,x)$, six ways to roll $(2,x)$, six ways to roll $(x,1)$, and six ways to roll $(x,2)$. This double counts the pairs $(1,1),(1,2),(2,1),(2,2)$. So there are $20$ ways to roll a $1$ or a $2$, meaning the odds of not doing so are $16/36=4/9$
A: I noticed after writing this that lulu already posted this exact solution
in a comment, so this is a community wiki response.

If $A$ is the event "roll $1$ on either of the two dice" and
$B$ is the event "roll $2$ on either of the two dice"
then your formula $1-P(A \text{ or } B)$ is correct.
But I think that $P(A)$ and $P(B)$ are each as hard to compute
as the original problem, so you might as well skip those steps.
I would suggest instead the following events:
$$ G = \text{the first die is neither $1$ nor $2$,}$$
$$ H = \text{the second die is neither $1$ nor $2$.}$$
Then the probability you want is $P(G \text{ and } H)$.
Since the dice roll independently,
$$P(G \text{ and } H) = P(G) \times P(H).$$
