27. There are 4 letters $A, B, C, D$ and 3 mailboxes numbered 1, 2, 3. If the 4 letters can be placed in any of the mailboxes, until all are placed. Find: (1) the probability that mailbox 3 has exactly 1 letter; (2) the probability that letter A is placed in mailbox 1 or 2.
Solve: Fill in $3 \sqrt{2}$. Reason: Let $f(\theta)=y$, since the variance of $\sqrt{1-\cos \theta+\sin \theta}, \sqrt{\cos \theta+2}, \sqrt{3-\sin \theta}$ is $s^{2}=\frac{1}{3}\left[(\sqrt{1-\cos \theta+\sin \theta})^{2}+(\sqrt{\cos \theta+2})^{2}+(\sqrt{3-\sin \theta})^{2}-\right.$ $\left.\frac{1}{3}(\sqrt{1-\cos \theta+\sin \theta}+\sqrt{\cos \theta+2}+\sqrt{3-\sin \theta})^{2}\right]=\frac{1}{3}\left(6-\frac{1}{3} y^{2}\right) \geqslant 0$, thus $0<y \leqslant$ $3 \sqrt{2}$, where the equality holds if and only if $\sqrt{1-\cos \theta+\sin \theta}=\sqrt{\cos \theta+2}=\sqrt{3-\sin \theta}$, i.e., $\theta=\frac{\pi}{2}$.
27. First, find the total number of basic events, by discussing the delivery of each letter step by step: The first step is to deliver the first letter, which has 3 different ways, the second, third, and fourth steps, delivering the remaining 3 letters each have 3 different ways, so $n=3 \times 3 \times 3 \times 3 \times 3=3^{4}$.
(1) This sub-question is divided into two steps: The first step is to choose 1 letter from 4 letters to be put into mailbox No. 3, which has $C_{1}^{1}$ ways, the second step is to put the remaining 3 letters into the remaining 2 mailboxes, which has $2^{3}$ ways, i.e.,
$m=C_{4}^{1} \cdot 2^{3}$, so $P=\frac{C_{4}^{1} \cdot 2^{3}}{3^{4}}=\frac{32}{81}$.
(2) The first step is to choose 1 mailbox from mailboxes No. 1 and No. 2 to put in letter $A$, which has $C_{2}^{1}$ ways, the second step is to put the remaining 3 letters into 3 mailboxes (different from sub-question (1)), which has $3^{3}$ ways, i.e.,
$m=C_{2}^{1} \cdot 3^{3}$, so $P=\frac{C_{2}^{1} \cdot 3^{3}}{3^{4}}=\frac{2}{3}$.