10. (14 points) Let $n$ be a given integer greater than 2. There are $n$ bags that are indistinguishable in appearance, the $k$-th bag ($k=1,2, \cdots, n$) contains $k$ red balls and $n-k$ white balls. After mixing these bags, one bag is randomly selected, and three balls are drawn consecutively from it (without replacement). Find the probability that the third ball drawn is white.
10. Suppose the selected bag is the $k$-th bag, the number of ways to draw balls three times in a row is $n(n-1)(n-2)$.

The color combinations of the three draws where the third ball is white are as follows:
(White, White, White) number of ways:
$$
(n-k)(n-k-1)(n-k-2) \text{, }
$$
(White, Red, White) number of ways:
$$
k(n-k)(n-k-1) \text{, }
$$
(Red, White, White) number of ways:
$$
k(n-k)(n-k-1) \text{, }
$$
(Red, Red, White) number of ways:
$$
k(k-1)(n-k) \text{. }
$$

Thus, the number of ways to draw a white ball on the third draw is
$$
\begin{array}{l}
(n-k)(n-k-1)(n-k-2)+ \\
k(n-k)(n-k-1)+ \\
k(n-k)(n-k-1)+k(k-1)(n-k) \\
=(n-1)(n-2)(n-k) .
\end{array}
$$

Therefore, the probability of drawing a white ball on the third draw from the $k$-th bag is $p_{k}=\frac{n-k}{n}$.

The probability of selecting the $k$-th bag is $\frac{1}{n}$, so the required probability is
$$
\begin{aligned}
p & =\sum_{k=1}^{n} p_{k} \cdot \frac{1}{n}=\sum_{k=1}^{n} \frac{n-k}{n} \cdot \frac{1}{n} \\
& =\frac{1}{n^{2}} \sum_{k=1}^{n}(n-k)=\frac{1}{n^{2}} \sum_{i=0}^{n-1} i=\frac{n-1}{2 n} .
\end{aligned}
$$