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Conditional Probability, Failure of Concrete Beam

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A concrete beam my fail by shear or flexure. The failure probability in shear is equal to the failure probability in flexure, and the probability of failure in shear when the beam is loaded beyond is flexure capacity (ie, it has already failed in flexure) is 80%. Find the failure probability of the beam in flexure, given that that the probability of failure of the beam is 0.2.

    2. Relevant equations

    P(A | B) = P(A [tex]\cap[/tex] B) / P(B) (conditional probability equation)

    3. The attempt at a solution

    So I started out by giving everything that's known:

    P(Failure) = 0.2 ==> P(Failure Shear [tex]\cup[/tex] Failure Flexure) = 0.2
    P(Failure Shear) = P(Failure Flexure)
    P(Failure Shear | Failure Flexure) = 0.8

    What I want to do is plug into the above equation, P(A | B) = P(A [tex]\cap[/tex] B) / P(B), since this seems to be a conditional probability problem. Plugging in:

    0.8 = 0.2 / P(Failure Flexure) -> P(Failure Flexure) = 0.25 = 25%

    However, I don't believe this is correct because I believe the 0.2 given is P(Failure Shear [tex]\cup[/tex] Failure Flexure), not P(Failure Shear [tex]\cap[/tex] Failure Flexure). I believe this because the 0.2 means probability of failure due to shear *or* flexure, which signifies union, not shear AND flexure, which signifies intersection.

    Could anybody steer me in the right direction? This is probably a very elementary problem but probability has never really 'clicked' with me. Thanks!
    Last edited: Sep 27, 2009
  2. jcsd
  3. Sep 27, 2009 #2
    I think I may have gotten an answer...but I'd like some confirmation of my work if you guys wouldn't mind.

    First off,
    S = Shear
    F = Flexure

    P(S [tex]\cup[/tex] F) = 0.2
    P(S | F) = 0.8

    P(S [tex]\cup[/tex] F) = P(S) + P(F) - P(S [tex]\cap[/tex] F)
    P(S [tex]\cup[/tex] F) = P(S) + P(F) - P(S | F)P(F) (conditional probability property)
    P(S [tex]\cup[/tex] F) + P(S | F)P(F) = P(S) + P(F)

    Dividing through by P(F) on the left,

    P(S [tex]\cup[/tex] F)/P(F) + P(S | F) = 1 + 1, since P(F)=P(S)

    Therefore, plugging in and solving:

    0.2/P(F) + 0.8 = 2

    Solving for P(F), I get 16.67%. Any checking of my work would be great. Thanks :).
    Last edited: Sep 27, 2009
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