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Conditional Probability Problem

  1. Jul 11, 2015 #1
    An electrical system consists of four components. The system works if components A and B work and either of the components C or D works. The reliability (probability of working) of each component is given by P(A) = 0.9, P(B) = 0.9, P(C) = 0.8, and P(D) = 0.8. Find the probability that (a) the entire system fails to work and (b) component C does not work, given that the entire system fails to work. Assume that the four components work independently.

    Thus far I have solved part a by solving for the probability that the entire system works, denoted P(E) where E is the event that the entire system works, then subtracting that from 1; this gave me the result P(E') = 0.2224 (where E' is the complement of E). I am just having trouble solving part b.

    I have found that P(C'|E') = [P(C')xP(E'|C')]/P(E') = P(C' n E')/P(E') where n is read "intersect". I really do not see where to go from here though. I have values for P(C') and P(E') but no values for P(E'|C') or P(C' n E'). Any help would be very much appreciated. Thank you in advance.
     
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  3. Jul 11, 2015 #2

    mfb

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    Which components have to (not) work if C works and the whole system does not work? What does that imply for P(C' n E')?
    The other unknown values can be analyzed in the same way.
     
  4. Jul 12, 2015 #3

    verty

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    I think [P(C')xP(E'|C')]/P(E') is easier to calculate than P(C' n E')/P(E') and probably this is always true when one uses Bayes' rule. In fact, I think it is probably a general principle to use ##P(A \cap B) = P(A | B) * P(B)## to get rid of any intersections so that it is easier.

    Ok, P(E'|C'). We are given that C has failed. There is no C, it is a short circuit. What is the probability of the system failing in this configuration?
     
    Last edited: Jul 12, 2015
  5. Jul 12, 2015 #4
    @verty I could see how that may be slightly easier. So would this be the correct way to go about it then?

    [P(C')xP(E'|C')]/P(E') = [P(C')x(1-P(E|C')]/P(E') = [P(C')x(1-A U B U C' U D)]/P(E') = [P(C')x(1-P(A)xP(B)xP(C')xP(D)]/P(E'). As I am new to this type of math, I do not know if the relation I used, P(E'|C') = (1-P(E|C')) would hold true or not.
     
  6. Jul 12, 2015 #5

    verty

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    This is doing my head in, all these symbols. Let's go back to square 1. You have an electrical system, you have components. You are trying to find how vulnerable the system is when C is removed or has failed. It's a real world problem and you are coming up with crazy formulas that don't seem to relate to anything real.

    No engineer is going to look at this question and start scribbling all of that junk down, now are they? They are going to say, hmm, P(C'|E'), that seems backwards, I can switch them by using Bayes' rule. Hmm, P(E'|C'), oh but that's obvious, that's how vulnerable the system is when I remove C.

    You need to stay focused on what you are trying to do. It should be simple, it should be straight-forward. And you need to think about what is really going on and not what is a bunch of symbol-twisting at the end of the day. This should be an easy question, and I know it's new for you but the truth is usually simple, that's what they always say. So look at this again and try not to make it too complicated, it just isn't worth getting into this forest of symbols when the problem should be an easy one.

    Okay? I hope you see now that actually there is an easier way to think about this and most problems can be handled in the same way, the KISS principle.
     
  7. Jul 12, 2015 #6

    haruspex

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    I assume you mean intersections, not unions, and you need a P() around the event combination.
    But why is there a C' In there? This is the expansion of P(E|C').
     
  8. Jul 14, 2015 #7

    RUber

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    For the first problem, you already found the probability that A and B are both working. This is a necessary condition for the system to work.
    If C is not working, what are the odds that D is working, that is the only way the whole system will work. Remember, these components, A, B, C, and D, are all independent, so you don't need to use conditional formulas, if you don't want to.
    1-P(E|C') = P(E'|C') since in this situation (C'), you only have E and E' as possible outcomes.
     
  9. Jul 15, 2015 #8
    For part (b), using "brute force engineering" I got an answer of 0.3165. Is this correct?
     
  10. Jul 15, 2015 #9

    RUber

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    What brute force are you using? Look back at post #3, given C' means you only have 3 possible functioning parts and all three have to work in order for the system to work.
    1- P(A&B&D)= 1-P(A) * P(B) * P(D), since they are independent.
     
  11. Jul 15, 2015 #10
    I simply calculated the probabilities for all 13 ways the system can fail and added them to get 0.2224.
    Then added together the subset that had C failing to get 0.0704.
    Then 0.0704/0.2224 = 0.3165.

    Is there another brute force way?
     
  12. Jul 15, 2015 #11

    RUber

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    It looks like you found P(C'|E') = P(C' & E') / P(E'). The question asks for P(E'|C') = P(C'&E')/P(C').
     
  13. Jul 15, 2015 #12
    Does not "(b) component C does not work, given that the entire system fails to work." ask for P(C'|E')? (Sorry, this is the first time I've seen this nomenclature.)
     
  14. Jul 15, 2015 #13

    RUber

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    Wow...I was reading it backward this whole time. Sorry about that. Yes...your brute force method was correct, and I was doing it wrong.
     
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