Question 194: Let $n$ be a given integer greater than 2, and there are $n$ bags that are identical in appearance. The $k(1 \leq k \leq$ $n)$-th bag 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.
Question 194, Solution: Let the selected bag be the $k$th bag, the number of ways to draw balls three times in succession is $n(n-1)(n-2)$. The color combinations for the three draws where the third ball is white are as follows:
(White, White, White) has the number of ways as $(n-k)(n-k-1)(n-k-2)$:
(White, Red, White) has the number of ways as $(n-k) k(n-k-1)$;
(Red, White, White) has the number of ways as $k(n-k)(n-k-1)$;
(Red, Red, White) has the number of ways as $k(k-1)(n-k)$;
Thus, the number of ways to draw a white ball on the third draw is:
$$
\begin{array}{l}
(n-k)(n-k-1)(n-k-2)+2(n-k) k(n-k-1)+k(k-1)(n-k) \\
=(n-1)(n-2)(n-k)
\end{array}
$$

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: $P=\sum_{k=1}^{n} \frac{1}{n} \cdot \frac{n-k}{n}=\frac{n-1}{2 n}$.