What is the probability of getting exactly 3 two's OR three's when a die is rolled 8 times? What is the probability of getting exactly $3$ two's OR three's when a die is rolled $8$ times?
I know that $P(E) = |E| / |S|$. 
I believe that $|S| = 36$, since there are $36$ different combinations when rolling a die. I am not sure how to get the probability of rolling exactly $3$ two's OR three's when rolling $8$ times.

Probability of exactly $3$ twos is $\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5$.
Probability of exactly $3$ threes is $\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5$.
Probability of exactly $3$ twos and exactly $3$ threes is $\binom{8}{3}\binom{5}{3}\left(\frac16\right)^6\left(\frac46\right)^2$
Thus, Inclusion-Exclusion says that the probability of exactly $3$ twos or $3$ threes is
$$
\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5+\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5-\binom{8}{3}\binom{5}{3}\left(\frac16\right)^6\left(\frac46\right)^2=\frac{7105}{34992}
$$
