When one throws three equal dice, how could we find what the probability that at least one die is holding the number 6 would be?
Here’s my solution. The sample space is $$\mathbb C = \{(x,y,z)\mid x,y,z, \in \mathbb N\setminus\{0\}, 1\le x,y,z \le 6 \}.$$
Then, the event space is $$C = \{(x,y,z) \in \mathbb C \mid (x,y,z) \in (6,y,z)\cup(x,6,z)\cup(x,y,6)\}.$$
Now, $|\mathbb C| = 6\cdot6\cdot6 = 216$ and $|C| = 36+36+36 - (6+6+6) + 1 =91$. Thus the probability equals $\frac{91}{216}$.
Is this correct?
This a good applicaiton of the inclusion-exclusion principle, and a correct solution. Depending on the level of detail required, it might be a good idea to explain briefly why $|C| = 3\cdot 36-3\cdot 6 + 1$, though (maybe even why $|\Bbb C| = 6^3$). If it has been covered, then simply referring to the inclusion-exclusion principle might be enough. If not, then writing down the general result before filling in the numbers will help: $$ |X\cup Y\cup Z| = |X| + |Y| + |Z| - |X\cap Y| - |X\cap Z| -|Y\cap Z| + |X\cap Y\cap Z| $$ In the end, how much detail is required is up to your teacher / tutor / whoever is correcting this, and not something that can be answered by a stranger on the internet.