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.
Best Answer
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 $$\frac{458752}{1679616} \approx 27.3\%.$$