The adjacent map is part of a city: the small rectangles are blocks, and the paths in between are streets. 
Each morning, a student walks from intersection $A$ to intersection $B$, always walking along streets shown, 
and always going east or south. For variety, at each intersection where he has a choice, he chooses with 
probability $\frac{1}{2}$ whether to go east or south. Find the probability that through any given morning, he goes through $C$. 

$\text{(A)}\frac{11}{32}\qquad \text{(B)}\frac{1}{2}\qquad \text{(C)}\frac{4}{7}\qquad \text{(D)}\frac{21}{32}\qquad \text{(E)}\frac{3}{4}$
Solution 1

The probability that the student passes through $C$ is the sum from $i=0$ to $3$ of the probabilities that he enters intersection $C_i$ in the adjoining figure and goes east. The number of paths from $A$ to $C_i$ is ${{2+i} \choose 2}$, because each such path has $2$ eastward block segments and they can occur in any order. The probability of taking any one of these paths to $C_i$ and then going east is $(\tfrac{1}{2})^{3+i}$ because there are $3+i$ intersections along the way (including $A$ and $C_i$) where an independent choice with probability $\tfrac{1}{2}$ is made. So the answer is \[\sum\limits_{i=0}^3 {{2+i} \choose 2} \left( \frac{1}{2} \right)^{3+i} = \frac{1}{8} + \frac{3}{16} + \frac{6}{32} + \frac{10}{64} = \boxed{\frac{21}{32}}.\]