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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

Figure 2.10

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.

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

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  • $\begingroup$ It may be useful to know how to simplify $B \cap (B\cup C)$. $\endgroup$ Sep 15 '16 at 0:04
  • $\begingroup$ is it. P(B∩(A∩(B∪C)∩D)) or P(B∩(B∪C))? are they the same thing? sorry im just really confuse in this kind of things $\endgroup$ Sep 15 '16 at 0:06
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    $\begingroup$ Thank you very much! This is my first Statistics class so im kinda confused. lmaoo but i really thank you! $\endgroup$ Sep 15 '16 at 0:31
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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

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