What is the probability of getting from point $A$ to point $B$ given each path has $1-p$ of being open? 
What is the probability of getting from point A to point B, given that the probability of each line segment being close is p and being open is (1-p) and each line segment is independent of the others?  
For this I used the inclusion-exclusion rule for a set of 4.
P1=Top path open
P2=Bottom path open
P3=Top left, middle, bottom right being open
P4=Bottom left, middle, top right being open
Seems reasonable but not 100% sure if it's the right approach?
 A: I will call "success" a path from A to B.
You will be successful every time there are four or five open paths.  There are exactly eight ways to have success with three open paths (the two top edges and any other edge, the two bottom edges and any other edge, and the two length-3 paths that include the middle edge).  Finally, there are exactly two successes with exactly two open paths.
Since the probability of a path being closed is $p$ and the five edges are all independent, the final probability of success is 
$$(1-p)^5+5p(1-p)^4+8p^2(1-p)^3+2p^3(1-p)^2$$
A: Here's the inclusion-exclusion calculation as a sanity check:
\begin{align}
&[2(1-p)^2+2(1-p)^3]-[5(1-p)^4+(1-p)^5]
+4(1-p)^5
-(1-p)^5 \\
&=1-2 p^2 - 2 p^3 + 5 p^4  - 2 p^5
\end{align}
But it doesn't match @MatthewDaly's $1-p- 2 p^3 + 3 p^4 - p^5$.  Both formulas correctly yield $1$ when $p=0$ and $0$ when $p=1$.
The discrepancy is because there are 16 successful arrangements, not 13.  If you correctly include [bcde/a], [acd/be], and [abe/cd], Matthew's approach yields: 
$$(1−p)^5+5p(1−p)^4+8p^2(1−p)^3+2p^3(1−p)^2 = 1-2 p^2 - 2 p^3 + 5 p^4  - 2 p^5,$$
which agrees with my inclusion-exclusion.
