Conditional probability in component network I am a little stuck on this problem involving the below reliability network:

The probability that any given component works is independent with probability $p$, and fails with probability $q=1-p$. I am trying to find the probability that components $1$ and $5$ work given that the system works. 
Starting with the probability the system works I get the probability that components $1$ and $2$ work is $p^2$, the probability that components $4$ and $5$ work is $1-q^2$. Therefore the probability that components $3$, $4$, and $5$ work is $p(1-q^2)$, hence:
$$P(\textrm{System Works})=1-(1-p^2)(1-p(1-q^2))$$
This can be simplified but I do not think it is too important. I then try to find the probability that $1$, $5$, and the system works. I am struggling on this part, it seems like there are many combinations where this condition is satisfied, so I was wondering if there is a better way to consider all the events which the system works as well as components $1$ and $5$?
Just to clarify what I mean by a reliability network. The probability that the system works is a path getting from the right to left side (or vice versa) passing through components that work with probability $p$. You cannot pass through a component if it fails (probability $1-p$ defined to be $q$).
 A: \begin{eqnarray*}
\mbox{Let } && S = \mbox{System works} \\
&& W = \mbox{1 and 5 both work.}
\end{eqnarray*}
With the obvious notation, and using distributivity rules:
\begin{eqnarray*}
S &=& (3 \cap (4 \cup 5)) \;\cup\; (1 \cap 2) \\
&=& (3 \cap 4) \;\cup\; (3 \cap 5) \;\cup\; (1 \cap 2). \\
W &=& 1 \cap 5. \\
\therefore S \cap W &=& \left[(3 \cap 4) \;\cup\; (3 \cap 5) \;\cup\; (1 \cap 2)\right] \;\cap\; (1 \cap 5) \\
&=& (1 \cap 3 \cap 4 \cap 5) \;\cup\; (1 \cap 3 \cap 5) \;\cup\; (1 \cap 2 \cap 5) \\
&=& (1 \cap 3 \cap 5) \;\cup\; (1 \cap 2 \cap 5). \\
\therefore P(S \cap W) &=& P(1 \cap 3 \cap 5) + P(1 \cap 2 \cap 5) - P(1 \cap 2 \cap 3 \cap 5) \quad\mbox{by inclusion-exclusion} \\
&=& 2p^3 - p^4.
\end{eqnarray*}
A: I would use chain rule for:
$P$(System works,One ok,Five ok)=$P$(System works|One ok,Five ok)$P$(One ok,Five ok)
Assuming one ok and five ok:


*

*one can be eliminated from the network (passes always through). 

*similarly you can eliminate five - there is just a straight line

*finally, if you pass five all the time, you can eliminate four: you can use always the line where was five

*the remaining nodes are two and three, in the same arrangement as four and five in the original picture


Therefore we get:
$P$(System works|One ok,Five ok) = $1-q^2$
From the formulation and independence, we get
$P$(One ok,Five ok) =  $p^2$
Thus $P$(System works,One ok,Five ok) = $p^2(1-q^2)$
