14・32 The figure below is a part of a city map. Small rectangles represent houses and buildings, and the spaces between them represent streets. There is a student who walks from point $A$ to point $B$ every morning along the streets shown in the figure, and he can only move east or south. At each intersection, he has a probability of $\frac{1}{2}$ to choose either east or south (each choice is independent of the others). The probability that the student passes through point $C$ is
(A) $\frac{11}{32}$.
(B) $\frac{1}{2}$.
(C) $\frac{4}{7}$.
(D) $\frac{21}{32}$.
(E) $\frac{3}{4}$.
[Solution] As shown in the figure, according to the walking rules specified in the problem, the student must pass through one of the points $C_{i}(i=0,1,2,3)$, and then move eastward. That is, the probability that the student passes through $C$ is equal to the sum of the probabilities of passing through each of the points $C_{i}$ and then moving eastward.

In the paths from $A$ to $C_{i}$, there are two eastward segments and $i$ southward segments, so there are a total of $2+i$ segments, with the $i$ southward segments appearing in any order.

Therefore, the number of paths from $A$ to $C_{i}$ is $C_{2+i}^{i}$ (of course, we can also consider that the two eastward segments can appear in any order, so the combination number is $C_{2+i}^{2}$, but $C_{2+i}^{i}=C_{2+i}^{2}$).

Moreover, at each intersection, the probability of moving east or south is $\frac{1}{2}$. Since each path from $A$ to $C_{i}$ passes through a total of $3+i$ intersections (including $A$ and $C_{i}$), the probability of moving east on this path is $\left(\frac{1}{2}\right)^{3+i}$. Therefore, the probability sought in the problem is
$$
\begin{aligned}
& \sum_{i=0}^{3} C_{2+i}^{i}\left(\frac{1}{2}\right)^{3+i} \\
= & \frac{1}{8}+\frac{3}{16}+\frac{6}{32}+\frac{10}{64} \\
= & \frac{21}{32} .
\end{aligned}
$$

Therefore, the answer is $(D)$.
Note If we write out the probability of passing through each intersection step by step, we can get the tree diagram on the upper right (where the final 1 is only for checking the calculation).

From the diagram, we can clearly see that the 20 paths from $A$ to $B$ passing through $C$ (there are a total of 35 paths from $A$ to $B$) are not equally likely.