A die is thrown 10 times. What is the probability that $6$ isn't registered and that at least one "1" is registered. 
A die is thrown 10 times. What is the probability that $6$ isn't registered and that at least one "1" is registered.

$A$, first occurrence - that $"6"$ isn't registered
$B$, second occurrence - that at least one $"1"$ is registered.
What I know: How to find $P(A)$ and $P(B)$ (over their complements)
What I'm having trouble with: Determining if I need P(A\mid B) or just P(AB) either why I do not know how to find P(AB). I was thinking total probability but got no farther than that...

Let $A$ be the event at least one $2$, and $B$ the event no $6$. We want $\Pr(A\cap B)$. This is $\Pr(A|B)\Pr(B)$.
It is not hard to see that $\Pr(B)=\frac{5^{10}}{6^{10}}$.
For $\Pr(A|B)$, we can proceed as follows. Given $B$, there are $5^{10}$ equally likely outcomes. Of these, $4^{10}$ have no $2$, so $5^{10}-4^{10}$ have at least one $2$. It follows that
$$\Pr(A|B)=\frac{5^{10}-4^{10}}{5^{10}}.$$
Now we can find $\Pr(A\cap B)$.
Another way: The probability of no $6$ is $\left(\frac{5}{6}\right)^{10}$. The probability of no $6$ and no $2$ is $\left(\frac{4}{6}\right)^{10}$. It follows that the probability of no $6$ and at least one $2$ is
$$\left(\frac{5}{6}\right)^{10}-\left(\frac{4}{6}\right)^{10}.$$
