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.
 A: Total number of possible rolls: $6^8 = 1679616$.
I see two possible interpretations of the question. In the following, I answer both of them.
Interpretation 1:
There are exactly 3 twos or exactly 3 threes
Number of rolls with exactly 3 twos: $\binom{8}{3}\cdot 5^5 = 175000$.
(Explanation: $\binom{8}{3}$ is the number of choices of the positions of the twos. The factor $5^5$ comes from the remaining $5$ dice, which have $5$ possibilities (every number except two) each.)
Number of rolls with exactly 3 threes: $\binom{8}{3}\cdot 5^5 = 175000$.
Number of rolls with exactly 3 twos AND exactly 3 threes: $\binom{8}{3}\binom{5}{3}\cdot 4^2 = 8960$.
So by the sieve formula, the number of rolls with exactly 3 twos OR exactly 3 threes is $175000 + 175000 - 8960 = 341040$.
Thus, the probability is
$$
\frac{341040}{1679616} \approx 20.3\%.
$$
Interpretation 2: Exactly 3 dice show the digit 2 or 3
The number of such rolls  is $\binom{8}{3}\cdot 2^3 \cdot 4^5 = 458752$.
So the probability is
$$\frac{458752}{1679616} \approx 27.3\%.$$
A: 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
$$
\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5+\binom{8}{3}\left(\frac16\right)^3\left(\frac56\right)^5-\binom{8}{3}\binom{5}{3}\left(\frac16\right)^6\left(\frac46\right)^2=\frac{7105}{34992}
$$
A: total possibilities: $6^8$.  
$3$ twos or threes: $4^5\times 2^3\times{8\choose 3}$ (possibilities for remaining $5$ times possibilities for $3$ $2s$/$3s$ times which $3$ consist of $2s$ and $3s$)
${4^5\times 2^3\times{8\choose 3}}\over{6^8}$
interpretation was for $3$ twos or threes e.g., $2,3,3...;3,3,3...;3,2,3...;...$
A: Strictly speaking this is a problem of binomial distribution. 
Let $X$ be the random variable denoting the number of $2$'s in $8$ throws of a dice and let $Y$ be the random variable denoting the number of $3$'s in $8$ throws of a dice.
Here, $X,Y \approx Bin(8,\frac{1}{6})$ 
Therefore $P(X=3)= (^8_3)(\frac{1}{6})^3(\frac{5}{6})^5$  and $P(Y=3)= (^8_3)(\frac{1}{6})^3(\frac{5}{6})^5$
