10 manuscripts are placed in 30 folders and each manuscript needs 3 folders. What is the probability that a randomly selected 6 folders do not contain any complete manuscript?
Here is my solution
Let $\mathcal{M_i}$ denote event of i-th manuscript being complete. Let us find $$\mathbb{P}(\bigcup_{i=1}^{10}\mathcal{M_i})$$
To do that me may find $\mathbb{P}(\mathcal{M_1})$ which denotes of 1st manuscript being complete that is cotaining 3 folders of 1st manuscrpt and 3 of any other folder; therefore $$\mathbb{P}(\mathcal{M_1}) = \frac{\binom{27}{3}}{\binom{30}{6}}$$ and by inclusion exclusion we know $\sum_{i=1}^{10}\mathbb{P}(\mathcal{M_i}) = \binom{10}{1} \mathbb{P}(\mathcal{M_1})$.
Following the same logic we know that 2 mansucripts can be complete if half of 6 folders belong to one manuscript and other half to another and there is only one such combination for each case;$$\sum_{i<j}\mathbb{P}(\mathcal{M_i}\cap\mathcal{M_j}) = \binom{10}{2}\frac{1}{\binom{30}{6}} = \frac{45}{\binom{30}{6}}$$; we cannot have 3 or more complete manuscripts by selecting only 6 folders thus by inclusion exclusion;
$$\mathbb{P}(\bigcup_{i=1}^{10}\mathcal{M_i}) = \frac{10\binom{27}{3}-45}{\binom{30}{6}} $$
and the needed result is $1 - \mathbb{P}(\bigcup_{i=1}^{10}\mathcal{M_i}) = 0.950814702$
Anything wrong in my logic?
Here's a solution I wrote before reading your solution. It seems to be similar to your own. I believe that your solution is correct.
Let $S_1$ be the collection of all sets of $6$ folders that contain the entire first manuscript. Let $S_i$ (for $i = 2,\ldots,10$) be defined similarly.
Then $S = \cup_{i=1}^{10} S_i$ is the collection of all sets of 6 folders that contain at least one entire manuscript.
From the inclusion-exclusion principle, \begin{align} |S| &= \sum_{i=1}^{10} | S_i | - \sum_{i < j} | S_i \cap S_j| \\ &= \sum_{i=1}^{10} \binom{27}{3} - \sum_{i < j} 1 \\ &= 10 \binom{27}{3} - 45. \end{align} The total number of sets of 6 folders is $\binom{30}{6}$. Thus, the probability that a random selection of $6$ folders contains at least one entire manuscript is $$ p = \frac{10 \binom{27}{3} - 45}{\binom{30}{6}}. $$ The probability that a random selection of 6 folders contains no entire manuscript is $1 - p$.