Undirected probability graph Given the following graph:

Find the probability P(A->D) Where $P_{AD}, P_{CD}..$ Is the probability of failure to proceed to the next node.
My attempt:
The probability to go from A to D is $(1-P_{AD})$, From A to C is $(1-P_{AC})(1-P_{CD})$ And from A to B to C is $(1-P_{AB})(1-P_{BC})(1-P_{CD})$
Problem is I know I have to add the probabilities, but I should also subtract them.
$(A \rightarrow D)=(1-P_{AD})+(1-(1-P_{AD})(1-P_{AC})(1-P_{CD})+(1-(1-(1-P_{AD})(1-P_{AC}
(1-P_{CD})(1-P_{AB})(1-P_{BC})(1-P_{CD})$
But This doesn't seem reasonable, I think the problem is how I am accounting for one road being "broken".
EDIT: The question is asking for the total probability of going to Node D from node A given that the probability of each edge is the probability that the given edge is broken.
The suggested answer is $ (1 − _{}) + _{}(1 − _{})[1 − _{} + _{} (1 − _{})(1 − _{} ) ] $
 A: "The probability of going from $A$ to $D$" suggests that you might be thinking about a random walk question - if you take random steps in the graph, what is the probability you get from $A$ to $D$. However, judging by the suggested answer, that's not what the question means.
The question must actually mean: what is the probability that (if each edge has the corresponding probability of being deleted) it is possible to get from $A$ to $D$? That is, what is the probability that they are in the same connected component?
To answer that question, we reason as follows:

*

*If edge $AD$ is present, then it's definitely possible to go from $A$ to $D$.


*If edge $AD$ is absent, then edge $CD$ must be present for it to be possible to go from $A$ to $D$. Even if it is, we get two subcases:
A. Edge $AC$ could be present; since we're already assuming edge $CD$ is present, that would be enough.
B. Or, edge $AC$ could be absent; then, the only way left is for edges $AB$ and $BC$ to both be present.
Now, we compute the probabilities. Case 1 happens with probability $1 - p_{AD}$. Case 2 happens with probability $p_{AD}(1 - p_{CD})$, which is a factor that applies to both subcases. From there, subcase 2A happens with probability $1 - p_{AC}$, while subcase 2B happens with probability $p_{AC}(1 - p_{AB})(1-p_{BC})$.
Combining these, we get the suggested answer:
$$
   \underbrace{(1 - p_{AD})}_{\text{Case 1}} + \underbrace{p_{AD}(1 - p_{AC})}_{\text{Case 2}}[\underbrace{1 - p_{AC}}_{\text{Case 2A}} + \underbrace{p_{AC} ( 1 - p_{AB}) (1 - p_{BC})}_{\text{Case 2B}}]. 
$$
