Given that the system works, compute the probability that component is working. Suppose the diagram of an electrical system is
as given in Figure 2.10. What is the probability that
the system works? Assume the components fail independently

I got the answer for the probability of all the system works and it is $0.8037$
but there is another question.
Given that the system works, compute the probability that component B is working. 
I am not sure about this one. But
is the calculation like this? 
$$
P(A \cup B \cup C \cup D)/0.8037 
$$
or 
only $P(A \cup B \cup D)/0.8037$ since we are only talking about B is working given that the system works.
 A: The system is: $\rm A\cap(B\cup C)\cap D$ with:   $\mathsf P(A)=0.95\\\mathsf P(B)=0.7\\\mathsf P(C)=0.8\\\mathsf P(D)=0.9$
When given that the system works, we know components $\rm A$ and $\rm D$ must do so, and that at least one of $\rm B$ or $\rm C$ does too.
So: $\mathsf P(B\mid A\cap(B\cup C)\cap D)=\mathsf P(B\mid B\cup C)$
The question is then: find $\mathsf P(B\mid B\cup C)$. 

Because component failures are independent: $$\begin{align}\mathsf P(B\mid A\cap (B\cup C)\cap D) ~=~& \dfrac{\mathsf P(B\cap A\cap (B\cup C)\cap D)}{\mathsf P(A\cap(B\cup C)\cap D)} \\=~& \dfrac{\mathsf P(A\cap B\cap D)}{\mathsf P(A\cap(B\cup C)\cap D)} \\=~& \dfrac{\mathsf P(A)~\mathsf P( B)~\mathsf P( D)}{\mathsf P(A)~\mathsf P(B\cup C)~\mathsf P(D)} \\=~& \dfrac{\mathsf P(B)}{\mathsf P(B\cup C)} &\bbox[ghostwhite]{\color{ghostwhite}{=~\dfrac{\mathsf P(B)}{\mathsf P(B)+\mathsf P(C)-\mathsf P(B)~\mathsf P(C)}}}\\=~&\dfrac{\mathsf P(B\cap(B\cup C))}{\mathsf P(B\cup C)} \\[2ex]\therefore \mathsf P(B\mid A\cap (B\cup C)\cap D) ~=~& \mathsf P(B\mid B\cup C)\end{align}$$
A: If components are in serial (e.g A & B), all must work in order for the system to work.
If components are in parallel (e.g B & C), the system works if any of the components work.
Segment 1: P(A) = 0.95
Segment 2: 1 - P(B') x P(C') = 1-0.3 x 0.2 = 0.94
Segment 3: P(D) = 0.9
Probability entire system works = Segment 1 x Segment 2 x Segment 3
Probability entire system works =  0.95 x 0.94 x 0.9 = 0.8037
