B1 A bug moves in the coordinate plane, starting at $(0,0)$. On the first turn, the bug moves one unit up, down, left, or right, each with equal probability. On subsequent turns the bug moves one unit up, down, left, or right, choosing with equal probability among the three directions other than that of its previous move. For example, if the first move was one unit up then the second move has to be either one unit down or one unit left or one unit right.

After four moves, what is the probability that the bug is at $(2,2)$ ?
Answer: $\frac{1}{54}$
Solution 1:
The bug must alternate between moving up and right. On the first move, the probability the bug moves either up or right is $\frac{1}{2}$. On each subsequent move, the bug has three choices (down, left, and the other move that was not the bug's previous move); the probability the bug continues towards $(2,2)$ is $\frac{1}{3}$. Thus, the total probability is $\frac{1}{2}\left(\frac{1}{3}\right)^{3}=\frac{1}{54}$.

Solution 2: The required probability is equal to the number of ways to get from $(0,0)$ to $(2,2)$ in four moves divided by the number of all possible four-move paths starting from $(0,0)$.

There are two ways to get from $(0,0)$ to $(2,2)$ : either URUR or RURU. There are $4 \times 3 \times 3 \times 3$ four-move paths starting from $(0,0)$. So, the probability is $\frac{2}{4 \times 3 \times 3 \times 3}=\frac{1}{54}$.