Sonya the frog chooses a point uniformly at random lying within the square
$[0, 6]$ $\times$ $[0, 6]$ in the coordinate plane and hops to that point. She then randomly
chooses a distance uniformly at random from $[0, 1]$ and a direction uniformly at
random from {north, south, east, west}. All of her choices are independent. She now
hops the distance in the chosen direction. What is the probability that she lands
outside the square?
$\textbf{(A) } \frac{1}{6} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{4} \qquad \textbf{(D) } \frac{1}{10} \qquad \textbf{(E) } \frac{1}{9}$
WLOG, we assume Sonya jumps $0.5$ units every time, since that is her expected value. 
If Sonya is within $0.5$ blocks of an edge, she can jump off the board. Let us examine the region that is at most $0.5$ blocks from exactly one edge. 
[asy] import graph;  Label f; xaxis(0,6,Ticks(f, 6.0, 0.5)); yaxis(0,6,Ticks(f, 6.0, 0.5));  draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); filldraw((0,0.5)--(0.5,0.5)--(0.5,5.5)--(0,5.5)--cycle,gray); filldraw((0.5,0)--(0.5,0.5)--(5.5,0.5)--(5.5,0)--cycle,gray); filldraw((6,0.5)--(5.5,0.5)--(5.5,5.5)--(6,5.5)--cycle,gray); filldraw((0.5,6)--(0.5,5.5)--(5.5,5.5)--(5.5,6)--cycle,gray); [/asy]
If Sonya starts in this region, she has a $\dfrac14$ chance of landing outside (there's exactly one direction she can hop to get out). The total area of this region is $4\cdot0.5\cdot5=10.$ For this region, Sonya has a $\dfrac14$ chance, so we multiply $10$ by $\dfrac14$ to get $2.5.$
If Sonya is in one of the corner squares, she can go two directions to get out, so she has a $\dfrac24=\dfrac12$ chance to get out. The total area is $0.5\cdot0.5\cdot4=1$, so this region yields $\dfrac12\cdot1=\dfrac12.$
Adding the two, we get $3$. This is out of $36$ square units of area, so our answer is thus $\boxed{\textbf{(B) }\tfrac{1}{12}}.$
~Technodoggo