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Probability problem(minmal paths sets and cuts sets

  1. Oct 13, 2012 #1
    The solution attempt in the attacment
     

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  2. jcsd
  3. Oct 13, 2012 #2
    thank you in advance
     
  4. Oct 13, 2012 #3

    haruspex

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    The same failure (e.g. of A) occurs at more than one place in your expression. Looks to me that your calculation treats these as independent, which would be wrong.
    Try breaking it up as:
    P[A fails] * P[circuit fails given that A fails] + P[A does not fail] * P[circuit fails given that A does not fail]
     
  5. Oct 13, 2012 #4
    First of all many thanks
    But there is something that i cant understand,which is my tutor told me we use minmal path sets for determine reliability and mimmal cut sets to determine failure
    and he implement this in the class method and it works however here does not work,
    So can i conclude that the method he gave me cant be use anywhere or there is something missing in my mind
    I attached the class example to have alook on it and understand what im saying
     

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  6. Oct 14, 2012 #5

    haruspex

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    The example in the PDF made it look easy because it was such a simple circuit.
    There is nothing wrong with the principle of cutsets/minimal paths; the problem is with evaluating the probability expression.
    In the OP, you have P[(Af & Bf) v (Af & Df) v (Cf & Df)]. This is awkward because you have the same event occurring in more than one place in the expression. In order to multiply probabilities, the events must be independent; in order to add them they must be disjoint (or you must correct for the double counting using the principle of inclusion/exclusion as is done in the PDF).
    In this more complicated case you can write
    P[(Af & Bf) v (Af & Df) v (Cf & Df)] = P[(Af & Bf) v (Af & Df) v (Cf & Df) | Af] * P[Af] + P[(Af & Bf) v (Af & Df) v (Cf & Df) | ~Af] *(1-P[Af])
    = P[Bf v Df v (Cf & Df)] * P[Af] + P[Cf & Df] *(1-P[Af])
    = P[Bf v Df] * P[Af] + P[Cf & Df] *(1-P[Af])
    From there it's straightforward.
     
  7. Oct 14, 2012 #6
    Please bare with me,if i understand you ,you said that i should use inclusion and exclusion principle,and i agree with this,the following attachment show that i use this method,but answers still different,
    Please be patient,because i need to understand the problem in my understanding
     

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  8. Oct 14, 2012 #7

    haruspex

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    The principle says you add up the probs of X1 and X2, for example, but because they're not mutually exclusive you have to subtract the probability of X1 & X2 (because you've counted that twice). If X1 and X2 are independent, the joint probability is the product of the two probabilities. But here, some of the terms you're treating that way are not independent (e.g.because they both refer to A). So it isn't simply X1 prob * X2 prob. That's why you get the wrong answer.
     
  9. Oct 14, 2012 #8
    Great ,i UNDERSATND NOW,big thanks to you
    So because they are independent i cant use this method , so what will be the solution for this problem???
    Many thanks again
     
  10. Oct 14, 2012 #9

    haruspex

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    You can still use the inclusion/exclusion principle, but you have to calculate the joint probabilities correctly. The problem is that they are NOT independent in this circuit.
    To sum up:
    Principle of inclusion/exclusion, quite general (two event case for simplicity):
    P[X v Y] = P[X] + P[Y] - P[X & Y]
    There are three special cases which make this simpler:
    If X and Y are mutually exclusive, P[X & Y] = 0
    If X and Y are independent, P[X & Y] = P[X]*P[Y]
    If X implies Y, P[X & Y] = P[X]
    For the circuit you have, none of these apply. Can you understand the method I used a couple of posts ago?
     
  11. Oct 15, 2012 #10
    Exactly it works man,your are brilliant and thank you
    May i ask you why you choose A and consider it as bridge(onetime works and other fails)
    Why spesfically A
    I think because A make the whole problem and it makes events non independent
    I'm i right!!!!
     
  12. Oct 15, 2012 #11

    haruspex

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    I chose A because it appeared frequently in the expression. Fixing its value was sure to simplify things quickly. Glad to have helped.
     
  13. Oct 20, 2012 #12
    Brother i was reading in a book and i saw an example regarding to what we were speaking about
    and here the events are not mutally exclusive , because you can see {A} event seveal times
    So why this method works here
    explain please
    IM lost
     

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  14. Oct 20, 2012 #13

    haruspex

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    He's doing the same thing a slightly different way.
    First, he adds up the four probabilities for the OR conditions. But then he has to subtract out the probs of the pairwise combinations of these. This is the inclusion/exclusion principle at work. Look at one example of these pairs: (A&B) & (A&C&E). He reduces this (behind the scenes) to (A&B&C&E) before computing the probability: qA*qB*qC*qE.
    In my method, I simplified the logic combination before applying inclusion/exclusion. It's a matter of whichever you're more comfortable with.
     
  15. Oct 22, 2012 #14
    Thank you very much,
    Unfortuantily now i am confused so please bar with me, i use exactly the book method to solve the problem i rise in the upper posts and the answer were differnent if u try to solve them usiing minmal bath sets and minmal cut sets as he did, i also inclusion and exclusion principle in my solution
    Seems strange
    PLEASE HELP ME i have a lot of coursework to be done and i have to understand this in order to be able to proceed
     
  16. Oct 22, 2012 #15

    haruspex

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    Please post your working for the two methods (or send it to me through the Forum's email service) and I'll see if I can spot what's going wrong.
     
  17. Oct 24, 2012 #16
    this the solution brother and i use the same exclusion and inclusion principle
    so please help me man and explain in details
    please
     

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  18. Oct 24, 2012 #17

    haruspex

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    OK, you're still making the mistake of calculating joint probability as though events are independent when they're not.
    P((A&D)v(A&C)v(B&D)) = P(X1vX2vX3) = P(X1) + P(X2) + P(X3) - P(X1&X2) - P(X2&X3) - P(X3&X1) + P(X1&X2&X3)
    So far so good. But then you turn P(X1&X2) into P(X1)*P(X2). You can't do that because X1 and X2 are not independent.
    P(X1&X2) = P((A&D)&(A&C)), which simplifies to P(A&D&C). Do you see why?
    So P(X1&X2) = P(A)*P(D)*P(C) = 0.729 (because A, C and D are independent).
    So with p = 0.9, P((A&D)v(A&C)v(B&D)) = 3*p^2 - p^3 - p^3 - p^4 + p^4 = 0.972
     
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