A bug crawls from $A$ along a grid. It never goes backward, it crawls towards all the other possible directions with equal probability. For example:
- At $A$, it may crawl to either $B$ or $D$ with a 50-50 chance
- At $E$ (coming from $D$), it may crawl to $B$, $F$, or $H$ with a $\frac{1}{3}$ chance each
- At $C$ (coming from $B$), it will crawl to $F$ for sure
The questions are, from $A$:
- What is the probability of it landing at $E$ in 2 steps?
- What is the probability of it landing at $F$ in 3 steps?
- What is the probability of it landing at $G$ in 4 steps?
This problem can be solved by the standard tree technique. The answers are $\boxed{\frac{1}{2}}$, $\boxed{\frac{5}{12}}$, and $\boxed{\frac{7}{72}}$, respectively.
- The bug can travel from $A$ to $E$ in $2$ steps along two paths:
Therefore, the probability is $$\frac{1}{2}\times \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} =\frac{1}{2}$$
The bug can move from $A$ to $F$ in $3$ steps along three paths.
Therefore, the probability is $$\frac{1}{2}\times \frac{1}{2}\times 1 + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{3}+ \frac{1}{2}\times \frac{1}{2}\times \frac{1}{3}=\frac{5}{12} $$
- Three paths can lead the bug from $A$ to $G$ in $4$ steps.
Therefore, the probability is $$\frac{1}{2}\times \frac{1}{2}\times \frac{1}{3}\times \frac{1}{3} + \frac{1}{2} \times \frac{1}{2} \times \frac{1}{3}\times \frac{1}{2}+ \frac{1}{2}\times \frac{1}{2}\times \frac{1}{3}\times \frac{1}{3}=\frac{7}{72}$$