10. (20 points) From the set
$$
S=\{1,2, \cdots, n\}\left(n \in \mathbf{N}_{+}, n \geqslant 2\right)
$$

two different subsets $P$ and $Q$ are taken successively. Find the probability of the following events:
(1) $P \nsubseteq Q$, and $Q \nsubseteq P$;
(2) $\operatorname{Card}(P \cap Q)=k(0 \leqslant k \leqslant n-1)$.
10. Given that the set $S$ has $2^{n}$ subsets, the number of ordered subset pairs $(P, Q)$ is $\mathrm{A}_{2^{n}}^{2}=2^{n}\left(2^{n}-1\right)$.
(1) Consider the complementary event of “$P \nsubseteq Q$, and $Q \nsubseteq P$”:
“$P \varsubsetneqq Q$ or $Q \varsubsetneqq P$.
If $P \varsubsetneqq Q$, let $\operatorname{Card}(Q)=i(1 \leqslant i \leqslant n)$. Then $Q$ has $\mathrm{C}_{n}^{i}$ ways to be chosen. And $P$ is a proper subset of $Q$, so $P$ has $2^{i}-1$ ways to be chosen. Therefore, the number of subset pairs $(P, Q)$ satisfying $P \varsubsetneqq Q$ is
$$
\sum_{i=1}^{n} \mathrm{C}_{n}^{i}\left(2^{i}-1\right)=\sum_{i=0}^{n} \mathrm{C}_{n}^{i} 2^{i}-\sum_{i=0}^{n} \mathrm{C}_{n}^{i}=3^{n}-2^{n} \text {. }
$$

By symmetry, the number of ways to choose $Q \varsubsetneqq P$ is also $3^{n}-2^{n}$.
Thus, the probability that $P \nsubseteq Q$, and $Q \nsubseteq P$ is
$$
1-\frac{2\left(3^{n}-2^{n}\right)}{2^{n}\left(2^{n}-1\right)}=1-\frac{3^{n}-2^{n}}{2^{n-1}\left(2^{n}-1\right)} \text {. }
$$
(2) The number of subsets of $S=\{1,2, \cdots, n\}$ that contain $n$ is $2^{n-1}$. Therefore, the event
$$
\operatorname{Card}(P \cap Q)=k(0 \leqslant k \leqslant n-1)
$$

is equivalent to choosing two subsets $P^{\prime} 、 Q^{\prime}$ from the $(n-k)$-element set $S^{\prime}=S \backslash(P \cap Q)$ such that
$$
P^{\prime} \cap Q^{\prime}=\varnothing \text {. }
$$

Let $\operatorname{Card}\left(P^{\prime}\right)=i(0 \leqslant i \leqslant k)$. Then $P^{\prime}$ has $\mathrm{C}_{k}^{i}$ ways to be chosen. Thus, $Q^{\prime} \subseteq \complement_{s^{\prime}} P^{\prime}$.
Therefore, $Q^{\prime}$ has $2^{k-i}$ ways to be chosen.
At this point, the subset pair $\left(P^{\prime}, Q^{\prime}\right)$ has $2^{k-1} \mathrm{C}_{k}^{i}$ ways to be chosen.
Hence, the number of subset pairs $\left(P^{\prime}, Q^{\prime}\right)$ satisfying $P^{\prime} \cap Q^{\prime}=\varnothing$ is
$$
\sum_{i=0}^{k} 2^{k-i} \mathrm{C}_{k}^{i}=3^{k} \text { (ways). }
$$

Therefore, the probability that $\operatorname{Card}(P \cap Q)=k(0 \leqslant k \leqslant n-1)$ is $\frac{3^{k}}{2^{n}\left(2^{n}-1\right)}$.