Probability that the signal will pass through the scheme. 
What is the probability that the signal will pass through the scheme?
Every relay is open with the probability of $p$ and closed with the probability of $q$.

And if we got a signal at the end what conditional probability do we have that E was open?

The only way I can see is to consider every possible path from the beginning to the end, but it is to bulky. Are there any other ways to do it?
 A: Consider two separate cases:
1) If $E$ is closed then the signal will pass through the scheme iff 
$$((\text{A is closed}) \lor (\text{C is closed})) \land ((\text{B is closed}) \lor (\text{D is closed})).$$
That is
$$\neg((\text{A is open}) \land (\text{C is open})) \land \neg((\text{B is open}) \land (\text{D is open}))$$
and the probability is $q\cdot(1-p^2)\cdot(1-p^2)$.
2) If $E$ is open then the signal will pass through the scheme iff 
$$((\text{A is closed}) \land (\text{B is closed})) \lor ((\text{C is closed}) \land (\text{D is closed})).$$
That is
$$\neg(\neg((\text{A is closed}) \land (\text{B is closed}))) \land \neg((\text{C is closed}) \land (\text{D is closed}))))$$
and the probability is $p\cdot(1-(1-q^2)\cdot(1-q^2))$.
Therefore the probability that the signal will pass through the scheme is 
$$q(1-p^2)^2+p(1-(1-q^2)^2)=1-2p^2-2p^3+5p^4-2p^5$$
where we set $q=1-p$. Note the result is $1/2$ for $p=1/2$ (see a plot HERE).
Are you able to answer to the second question?
A: The following table shows the output depending on the inputs and the probabilities belonging to the actual gate combination. ($1$ for closed, $0$ for open.)

The sum of the probabilities belonging to the green area is the probability that the signal gets through.
The sum of the probabilities in the black-framed area ($P$) is the probability that the signal gets through and the E gate is open. 
The sum of the probabilities in the red-framed area ($Q$) is the probability that the E gate is open.
$\frac PQ$ is the conditional probability we are seeking.
