If the probabilities of the three components $A$, $B$, $C$ were independent of each other (in other words: uncorrelated), then it would indeed be correct to simply multiply the probabilities associated with $ABC$, $AB$, $AC$ and $BC$ working. Followed by adding up these four numbers, giving you the chance that least two components work.
However, as clearly stated in the text, the probabilities for components $B$ and $C$ are interdependent! This means that, before you can do a calculation as in the previous paragraph, you must first explore fully the interdependence of these probabilities.
Okay, let us do this. It is given that $C$ has a failure chance of $0.01$. Furthermore it is given that if $C$ fails, then the chance that $B$ fails is $0.55%$. This leads us to the conclusion that the probability that both $C$ and $B$ fail is equal to $0.01 * 0.55 = 0.0055$.
With a bit more effort we can derive that the probability that $C$ fails and $B$ works is equal to $0.0045$. The chance that $C$ works and $B$ fails is $0.1455$. The chance that both $C$ and $B$ work is $0.8455$. We now have the four values that determine the interdependence of $C$ and $B$.
Finally we can compute the overall probability that the system works. We get:
$$P = P(ABC) + P(AC) + P(AB) + P(BC)$$
$$P = 0.7 * (0.8455 + 0.1445 + 0.0045) + 0.3 * 0.8455 = 0.9498$$
Yes , it is a union . The current will flow when at least one of the relays will pass it through .
However , you do not simply add because the events are not disjoint . This then requires using the Principle of Inclusion and Exclusion (PIE) to avoid over-counting common outcomes .
$$\begin{align}\mathsf P(E_1\cup E_2\cup E_3) ~=~& {\mathsf P(E_1)+\mathsf P(E_2)+\mathsf P(E_3) \\[1ex]-\mathsf P(E_1\cap E_2)-\mathsf P(E_1\cap E_3)-\mathsf P(E_2\cap E_3)\\+\mathsf P(E_1\cap E_2\cap E_3)} \\[1ex]=~& 3(0.9)-3(0.9)^2+(0.9)^3\\=~& 0.999\end{align}$$
Since the events are independent , then the probabilities of the unions are the product of the probabilities of the events.
However the answer is somewhat easier to obtain by using complements . The current will not flow only when all of the relays block it .
$$\begin{align}\mathsf P(E_1\cup E_2\cup E_3) ~=~& 1-\mathsf P(E_1^\complement\cap E_2^\complement\cap E_3^\complement) &~&\text{de Morgan's Rules}\\[1ex] =~& 1-\mathsf P(E_1^\complement)\,\mathsf P(E_2^\complement)\,\mathsf P( E_3^\complement) && \text{Independence} \\=~& 1-(1-0.9)^3 \\[1ex]=~& 0.999 \end{align}$$
That is all.
$\blacksquare$
Best Answer
In my answer to your previous question, we found that
$$p(w)=\frac{51}{128}$$ where $w$ is the event "the current flows" in the given circuit.
By using the same approach, if $c$ works then $$p((a\lor b)\land c\land d)=\left(1-\left(\frac{1}{2}\right)^2\right)\cdot 1\cdot \frac{1}{2}=\frac{3}{8},$$ and therefore $$p(w|c)=\frac{1}{2}\cdot\left(1-\left(1-\frac{3}{8}\right)\cdot \left(1-\frac{1}{2}\right)\cdot \left(1-\frac{1}{2}\right)\right)=\frac{1}{2}\cdot\left(1-\frac{5}{32}\right)=\frac{27}{64}.$$ Hence, it follows that $$p(c|w)=\frac{p(c)\cdot p(w|c)}{p(w)}=\frac{\frac{1}{2}\cdot \frac{27}{64}}{\frac{51}{128}}=\frac{27}{51}\approx0.529$$ which is a bit greater then $p(c)=0.5$.