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Assessing probabilities for very rare events 291<br />

Central<br />

generator<br />

overheats<br />

0.2<br />

0.9<br />

Secondary cooling<br />

system works<br />

0.8<br />

No<br />

overheating<br />

OK<br />

Automatic shutdown<br />

mechanism fails<br />

0.05<br />

Catastrophic failure<br />

Prob = 0.2 × 0.1 × 0.01 × 0.05<br />

= 0.00001<br />

Secondary cooling<br />

system fails<br />

0.1<br />

Warning system<br />

fails<br />

0.01<br />

0.99<br />

Warning<br />

system works<br />

0.95<br />

Automatic<br />

shutdown works OK<br />

Automatic shutdown Catastrophic failure<br />

mechanism<br />

fails Prob = 0.2 × 0.1 × 0.99 × 0.02 × 0.05<br />

0.05<br />

Manual shutdown<br />

= 0.0000198<br />

fails<br />

0.95<br />

0.02<br />

Automatic<br />

0.98<br />

Manual<br />

shutdown works OK<br />

OK<br />

shutdown works OK<br />

Figure 10.4 – An event tree<br />

∴ p(catastrophic failure) = 0.00001 + 0.0000198<br />

= 0.0000298<br />

to assess the probability that each factor will contribute to it. Then, by<br />

using the multiplication and addition rules of probability, the overall<br />

probability of failure can be calculated (note that, for simplicity, here<br />

it has been assumed that all the factors operate independently). Of<br />

course, it is important to try to ensure that the tree is as complete<br />

as possible so that all possible causes of the event in question are<br />

included (see Chapter 6). However, it may be difficult or impossible to<br />

assess subjective probabilities for such events as human error, deliberate<br />

sabotage, and, by definition, unforeseen weaknesses in a system.<br />

Fault trees<br />

Sometimes it is easier to consider the problem from a different point<br />

of view. In contrast to event trees, fault trees start with the failure or<br />

accident and then depict the possible causes of that failure. For example,<br />

suppose that a decision maker wants to assess the probability that a<br />

vital pipeline in a chemical plant will fracture within the next 12 months.<br />

Figure 10.5 shows a fault tree for his problem. He considers that a<br />

fracture would occur if there is either a defective weld or excess pressure<br />

in the pipeline. Because either event on its own would be sufficient<br />

to cause the fracture these two events are connected to the ‘pipeline<br />

fracture’ node by an ‘or’ symbol. The excess pressure in turn, however,<br />

would only occur if there was both a regulator failure and a failure in the<br />

safety valve, so an ‘and’ symbol connects these two events to the ‘excess<br />

pressure’ node.

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