I just want to make sure that the scenario I'm describing is indeed a Bernoulli trial type of experiment. Here is the set up and query regarding the set up: A six-sided die is rolled 10 times. What is the probability that a 1 or a 6 is rolled exactly two times? (This is not to be understood as a 6 being rolled twice or a 1 being rolled twice, but either of the two being rolled twice, e.g. 1...1..., 1...6..., 6...1..., 6...6...). We can define the event $A=\{ 1,6 \}$ and $A^{c}= \{ 2,3,4,5 \}$.
I'm inclined to see this as a Bernoulli trial with $p= P(A)=1/3$ and $n=10$, and the answer being $\binom{10}{2}p^{2}(1-p)^{8}$. One just can think of an analogous scenario of selecting 10 balls (with replacement) from an urn containing 2 black marbles and 4 white marbles (where we identify 1 and 6 with black marbles and other numbers with white marbles), and asking for the probability of selecting 3 black marbles among the 10 chosen.
It may seem odd that we're effectively counting the four classes (each class can occur $4^{8}$ ways) of events, e.g. $1---6-----$ and $6---1-----$ and $1---1-----$ and $6---6-----$ as a single Bernoulli outcome, i.e. $SFFFSFFFFF$, but I'm almost (hence the question) sure that this still satisfies the criteria for a Bernoulli trial type of experiment, since the entire sample space is "flattened" in an analogous manner.