If you were working in an industry with significant risks, which of the risk met
ID: 3175254 • Letter: I
Question
If you were working in an industry with significant risks, which of the risk methods would you employ during the design stage of a new Plant? Explain the reasoning in your choices - which methods have you chosen and why have you chosen them and which methods have you not chosen and why did you not choose them? What is a non-coherent fault tree? How does one go about obtaining a cut set for a non-coherent fault tree? Explain the term failure rate and its significance for assessment and risk management. A water tower is located in an risk When an occurs, the active earthquake occurs, the probability that the tower will fail depends on the magnitude of the earthquake and also on the amount of storage in the tank at the time of shaking of the ground. For simplicity, assume that the tank is either full (F) half full (H) with relative likelihoods of 1 to 3. The earthquake magnitude may be assumed to be either strong (s) or weak (w) with relative frequencies 1 to 9. When a strong earthquake occurs, the tower will definitely collapse regardless of the storage level. However, the tower will certainly survive a weak earthquake if the tank is only half full. If the tank is full during a weak earthquake, it will have a 50/50 chance of survival. If the tower collapsed during a recent earthquake, what is the probability that the tank was full at the time of the earthquake?Explanation / Answer
Define, F = Full tower, H = Half-full tower, C = tower collapse, S = Strong earthquake,
W = Weak earthquake
Using the relative likelihood given,
Pr(F) = 0.25, Pr(H) = 0.75
Pr(S) = 0.1, Pr(W) = 0.9
Pr(C|S F) = Pr(C|S H) = 1 (Prob. of collapse given strong earthquake and full or half full tower)
Pr(C|W F) = 0.5 (Prob. of collapse given weak earthquake and full tower)
Pr(C|W H) = 0 (Prob. of collapse given weak earthquake and half-full tower)
Let’s calculate Probability of Collapse using total probability theorem
P(C) = Pr(C S H) + Pr(C S F) + Pr(C W H) + PR(C W F)
= Pr(C|S H)*Pr(S H) + Pr(C|S F) * Pr(S F) + Pr(C|W H) * Pr(W H) + Pr(C|W F) * Pr(W F)
Assume earth quake and quantity of water are statistically independent
Pr(C) = 1*0.1*0.75 + 1*0.1*0.25 + 0*0.9*0.75 + 0.5*0.9*0.25
= 0.2125
The probability that the tower is full, assuming tower collapse,
P(F|C) = Pr(F S|C) + Pr(F W|C)
= (Pr(C|F S)*Pr(F S))/(Pr(C))+(Pr(C|F W)Pr(F W))/(Pr(C))
= (1*0.25*0.1)/0.2125+ (0.5*0.25*0.9)/0.2125
= 0.647 = 64.7%