Find the number of paths given the probabilities of each move (Probability and Permutation) A robot is programmed to move on a flat surface one step at a time, either upward (U) or downward (D) or to the left (L) or to the right (R). Each move is independent of the preceding move and the associated probabilities of moving in the directions U, D, L and R are $0.2, 0.1, 0.1$ and $0.6$ respectively.

The robot starts from the origin O.
(i) Show that the probability that the robot lands at O after $2$ moves is $0.16$
(ii) The robot lands at O after $4$ moves. Find the number of paths that the robot can take (One such path is DUDU)
(iii) Hence find the probability that the robot lands at O after $4$ moves
(iv) What is the probability that the robot lands at point A after $4$ moves, given that it landed at A after the first $2$ moves?
My attempt:
(i) To return to original position, the $2$ moves must be opposite in direction. So, $0.1 \times 0.6 + 0.1 \times 0.2 + 0.6 \times 0.1 + 0.2 \times 0.1 = 0.16$
(ii) Is the answer $36$? I fix the first direction, say L. Then there are four possibilities, they are U,L,D,R. For each second direction, I enumerate them. So for a fix first direction, I obtain $9$ paths. Then $9 \times 4 = 36$.
(iii) Cannot start as I not sure whether my answer for part (ii) is correct or not.
(iv) $0.2 \times 0.6 \times 0.16 + 0.2 \times 0.6 \times 0.16 = 0.0384$
Are my answers correct?
 A: You answer to #1 looks fine. Your count for #2 does not looks correct, but you got the right answer. To solve #2 and #3, you can proceed as follows.
(ii) Do casework based on the number of initial directions.
Case 1: Two directions total (example: $LLRR$)
Pick one direction from the $4$ available. The other direction is automatically selected (it is the opposite of your first selection). Choose two of the four available spots to place your chosen direction. Then, divide by $2$ to adjust for overcounting ($LLRR$ could have been formed with chosen direction either $L$ or $R$). The count is:
$$\frac{4 * \dbinom{4}{2}}{2} = 12.$$
Case 2: Four directions total (example: $DLUR$)
Just arrange the four directions in order into four possible positions. The number of ways to do this is simply $4! = 24.$
The answer is $\boxed{36}.$
(iii) We will approach this probability calculation in exactly the same way as we did in part (ii). However, a more detailed analysis is needed for Case 1.
Case 1: Two directions total (example: $LLRR$)
If we choose left and right only, then there are only $6$ ways to arrange. We now use the probabilities and find that this contributes $(0.6)^{2} \times (0.1)^{2} \times 6 = \frac{27}{1250}.$
If we choose down and up only, then there are also only $6$ arrangements. Like in the previous calculation, this contributes $(0.2)^{2} \times (0.1)^{2} \times 6 = \frac{3}{1250}.$
This entire case contributes $\frac{3}{125}.$
Case 2: Four directions total
This is a simple count, and we see that the probability is $(0.1)(0.1)(0.2)(0.6) \times 24 = \frac{18}{625}.$
The answer is $\frac{3}{125} + \frac{18}{625} = \boxed{\frac{33}{625}}.$
To address part #4, you need to understand conditional probability. Notice that if you are already back at the starting point after $2$ moves, the probability that you will land back there after $2$ more moves is still the same. The answer, like it's given in part (i), is $\boxed{0.16}$
Hope this helped.
