A simple random walk given as $S(n) = 0,1, 2, 3, ..., n$ representing the sum of i.i.d random variables, with the following:
$$P(Y_i = -1)=p, P(Y_i = 0)=1-p-q, P(Y_i=1)=q$$
Find $P[S(3)=0]$ when $p=0.15, q = 0.45$.
Given that we have a sum of i.i.d random variables, then $P(Y_3 = 0) = (1-p-q)^3=(1-0.15-0.45)^3=0.40^3$
First time working with random walks, I'm just basing this based on the i.i.d random variables. Have I correctly approached this?
We find that there are 9 combinations
$$\begin{align}(0,0,0) &= 0.40^3 \\ (0,1,0) &= 0.40^2\cdot0.15 \\(1,0,0) &= 0.40^2\cdot0.15 \\(1,1,0) &= 0.40\cdot0.15^2 \\(0,-1,0) &= 0.40^2\cdot0.45 \\(-1,0,0) &= 0.40^2\cdot0.45 \\(-1,-1,0) &= 0.40\cdot0.45^2 \\(1,-1,0) &= 0.40\cdot0.15\cdot0.45 \\(-1,1,0) &= 0.40\cdot0.15\cdot0.45 \\\end{align}$$
$$P[S(3)=0]=0.4^3+0.4^2\cdot \:0.15+0.4^2\cdot \:0.15+0.4\cdot \:0.15^2+0.4^2\cdot \:0.45+0.4^2\cdot \:0.45+0.4\cdot \:0.45^2+0.4\cdot \:0.15\cdot \:0.45+0.4\cdot \:0.15\cdot \:0.45 = 0.40$$
Expanding my comments into a partial answer:
First, it looks like you're making an important mistake: I think you're conflating $S(3)$, which is the position of the random walk after 3 steps, and $Y_3$, which is the third step itself.
In your amended question, you worked hard to compute $\mathbb P(Y_3 = 0)$. I can confirm that you have correctly computed the probability that the third step is $0$, because you have exhausted all possibilities for how that might occur. Note that $\mathbb P(Y_3 = 0)$ is given to be $1 - p - q = 1 - 0.15 - 0.45 = 0.40$. So, the good news is that your calculation is correct -- but the bad news is that it's not useful, because it's not what you were asked to find (and it's a consequence of things you already knew without the laborious computation).
Instead, you were asked to find the probability that the walk was at position $0$ after three steps. This happens in three of the paths that you evaluated above -- specifically, $(1, -1, 0)$, $(-1, 1, 0)$, and $(0, 0, 0)$. However, there are others that you didn't consider that are relevant, such as $(1, 0, -1)$. Your task is to find all the triples that sum to $0$.
This can be daunting at first, but it might be easier if you group similar cases together. For instance, you already found the probabilities of $(1, -1, 0)$ and $(-1, 1, 0)$ to be the same; note that this is also the same as the probability of $(1, 0, -1)$, as well as several other paths that are unstated so far. So, you might consider grouping them by type and then counting how many of them there are.
Regarding your latest question in the comments: yes, directly counting the cases (as you've started to do) becomes considerably more complicated when you proceed to 4 steps. However, if you consider the grouping approach I outlined, the types don't get too much worse than they already are.