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Math Help - Probability Problem

  1. #1
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    Probability Problem

    A circuit has 3 components: A,B,C. Both B and C serve the same purpose and it is only necessary that one is operational. But A is distinct and must always work. Given that:

    Probability (A fails) = 0.8
    Probability (B fails) = 0.4
    Probability (C fails) = 0.8

    What is the overall probability of failure assuming these are all independent events?
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  2. #2
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    Quote Originally Posted by mr_motivator View Post
    A circuit has 3 components: A,B,C. Both B and C serve the same purpose and it is only necessary that one is operational. But A is distinct and must always work. Given that:

    Probability (A fails) = 0.8
    Probability (B fails) = 0.4
    Probability (C fails) = 0.8

    What is the overall probability of failure assuming these are all independent events?
    Do you mean failure of circuit? Then you need A' or A B' C'. I get 0.8 + (0.2)(0.4)(0.8).
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  3. #3
    MHF Contributor matheagle's Avatar
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    I found the complement, mainly because I didn't read the final question.
    Let A denote A suceeds....
    So in order to suceed we want.
    P(A(BUC))=P(A)P(BUC)=P(A)\bigl(P(B)+P(C)-P(BC)\bigr)
    =(.2)(.6+.2-(.6)(.2))=.136
    So 1-.136=.864 is correct.
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  4. #4
    Member u2_wa's Avatar
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    First map how can circuit fails:
    If A fails = 0.8
    If A works but B & C fails =0.2*0.4*0.8 =0.064

    Probability that circuit fails = 0.8+0.064<br />
=0.864
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