Probability in network reliability 
The probability that a link is working fine is given by $2/3$. 
a)Find the probability that there exists a path from A to B along which no link has failed. (Give a numerical answer.)
b) Given that exactly one link in the network has failed, find the probability that there exists a path from A to B along which no link has failed.
The first part is pretty straight-forward and I have solved it correctly. But I am pretty confused with my approach in the second part. I wanted to do systematically.
So here's is my approach: 
$S =$ there is a part from $A$  to $B$, $F_1 =$ link $1$ has failed,
$W_1 =$ link $1$ is working
, $F =$  exactly one link has failed.
So the probability that it is working is given by 
$P(S | F) = \frac{P(S \cap F)}{P(F)} $
So here $F = (F_1 \cap W_2 \cap W_3 \cap W_4 \cap W_5 ) \cup (W_1 \cap F_2 \cap W_3 \cap W_4 \cap W_5) \cup ... $
From here I am kind of lost. I don't understand where to go from here. Can anyone explain me further from here, how to proceed.
 A: Approach this problem in the "engineering" way.   
The links 1 and 2 are in series: that means that the whole works if both links are ok : $P_{1,2}=P_{1}P_{2}=4/9$. Same for links 3 and 4.
$L_{1,2}$ and $L_{3,4}$ are in parallel: the whole does not work if both paths are out :
$$1-P_{1,4}=(1-P_{1,2})(1-P{3,4})=25/81 \quad \to \quad P_{1,4}=56/81$$ .
Finally $L_{1,4}$ and $L_{5}$ are in series, so
$$P_{1,5}=P_{1,4}P_{5}=56/81 \cdot 2/3= 112/243$$
This is the answer to a) because if there is a path with no link in failure then the connection is working.
For question b), there are five possibilities to have exactly one link failed, all with the same probability ($1/5$). So
1/5 (L5 not working $\to\;P_{1,5}=0$)+4/5(one of the others failed $\to\;P_{1,5}=1$) $=4/5$
A: You are about to get to the answer. You got that
$$
F=(F1∩W2∩W3∩W4∩W5)∪(W1∩F2∩W3∩W4∩W5)∪...
$$
There are 5 terms in the above equation. Each has probability of $p*(1-p)^4$. Out of those 5 terms, in only 1 of them the path doesn't exist from A to B- that is if link 5 is the failed node. 
so $$P = 4*p*(1-p)^4 / 5*p*(1-p)^4 = 4/5$$
