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.


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.