The lost bug

Since each moves it has 5 choices, there are $5^3 = 125$ ways it can pick to move in 30 seconds. Only 10 of them lead to the opposite vertex. Therefore the probability is $10/125$ or $2/25$ .

It does not matter which direction the bug takes first. There is a probability of $\tfrac25$ that the bug's next move will take it to the ring of $5$ vertices adjacent to the opposite vertex, and hence make reaching the opposite vertex in $3$ moves possible, and then a probability of $\tfrac15$ that the bug's last move will finish the trip. Thus the probability of the bug getting to the opposite vertex is $\tfrac{2}{25}$ .

No. of ways to get to opp vertex: 5 × 2 × 1 =10 Total trips possible: 5 × 5 × 5 =125 Probability = 10/125 = 2/25

It can make 3 moves in 30 seconds. There are always 5 possible vertices it can move to at any given vertices on the icosahedron. Therefore there are 5^3 = 125 different routes. Out of these routes only 10 will lead to the opposite corner. There are five ways for the first edge, then two for the next and only one for the last move. This leads to 5 2 1 = 10 possible routes. 10/125 simplifies to 2/25. 2 + 25 = 27.