If you throw a dice $10$ times, what is the probability that there is at least one $1$ and at least one $6$? 
If you throw a dice $10$ times, what is the probability that there is
at least one $1$ and at least one 6?

My Idea:
An analogous question one could ask oneself is: Suppose, you have a combination lock with $n$ different cogwheels. For each cogwheel one has $6$ possible numbers.
So in total we have $6^n$ ways.
If we negate "at least one 1 and at least one 6", we get "no $1$ or no $6$ or neither a $1$ or a $6$" Wouldn't this just be:$$1-\frac{5^n+5^n+4^n}{6^n}$$?
 A: $$\begin{align}
P[\text{at least one } 1 \cap \text{at least one } 6] &= 1-P[\text{no }1 \cup \text{no }6] \\
\\
&= 1 - \left( P[\text{no }1] + P[\text{no }6] - P[\text{no }1 \cap \text{no }6]  \right) \\
\\
&= 1 - \frac{5^n+5^n-4^n}{6^n}
\end{align}$$
A: So you're right there are $5^n$ possibilities when you take one out and there are $4^n$ possibilities when you take both out. However there is some overlap. Some of those $5^n$ cases with "no $1$" also have "no $6$", which means you're over counting. Essentially you're counting "no $1$ and no $6$" three times, so you need to take two out, which should give:
$$
1 - \frac{2\cdot5^n - 4^n}{6^n}
$$
There's a theorem for this in probability that's slipping my mind right now, but I'll look for it.
Edit: Yep here is: the addition law of probability. In the second part you're trying to find the probability that there are "no $1$'s" or "no $6$'s":
$$
P(A\cup B) = P(A) + P(B) - P(A \cap B)
$$
In this case $P(A) = P(B) = \frac{5^n}{6^n}$ and the probability of neither (i.e. the intersection) is $P(A\cap B) = \frac{4^n}{6^n}$ which gives what we deduced above.
