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.
Best Answer
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: $$\begin{align}\mathsf P(B\mid A\cap (B\cup C)\cap D) ~=~& \dfrac{\mathsf P(B\cap A\cap (B\cup C)\cap D)}{\mathsf P(A\cap(B\cup C)\cap D)} \\=~& \dfrac{\mathsf P(A\cap B\cap D)}{\mathsf P(A\cap(B\cup C)\cap D)} \\=~& \dfrac{\mathsf P(A)~\mathsf P( B)~\mathsf P( D)}{\mathsf P(A)~\mathsf P(B\cup C)~\mathsf P(D)} \\=~& \dfrac{\mathsf P(B)}{\mathsf P(B\cup C)} &\bbox[ghostwhite]{\color{ghostwhite}{=~\dfrac{\mathsf P(B)}{\mathsf P(B)+\mathsf P(C)-\mathsf P(B)~\mathsf P(C)}}}\\=~&\dfrac{\mathsf P(B\cap(B\cup C))}{\mathsf P(B\cup C)} \\[2ex]\therefore \mathsf P(B\mid A\cap (B\cup C)\cap D) ~=~& \mathsf P(B\mid B\cup C)\end{align}$$