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Probability assessment

by Lucky mkhonza
Tags: assessment, probability
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Lucky mkhonza
#1
Nov5-07, 01:20 AM
P: 14
Dear folks,

Please help me in assessing the following scenario. I have a cooling circuit arranged in series: (1) Heat exchanger (probability = 10^(-4)/requirement), (2) Pump (probability = 10^(-3)/requirement), (3) Heat exchanger (probability = 10^(-3)/requirement) and connection of pipes (probability = 10^(-4)/requirement) (name the pipes as 4th component). In addition there is 5th component supplying energy to the pump (probability = 10^(-3)/requirement).

What is the probability that the cooling circuit corresponding to the scenario arranged above fails?

Thank you in advance
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GTrax
#2
Nov10-07, 05:49 AM
P: 127
Unlike choosing two winning numbers in a row, where the probabilities are simply multiplied, in your case, a failure of any one of the services makes failure more likely. You add them all up somehow, but the result must not exceed 1.0, however many are the ingredients, or how likely they are.
GTrax
#3
Nov13-07, 04:53 PM
P: 127
OK - I do have the answer. It comes from my son, who is is in his Master's degree Math finals.
We put the problem to him slightly differently, but you will understand. Your example is likely the simplest case, where the probabilities of failure you set out are independent.

Please be aware that if you are calculating for a real situation, then the values for any component may well be modified by an fail event in one of the others. For example, a pump breakdown might be a minor inconvenience, but the collateral effect on the other probabilities might modify them dramatically to swiftly alter the overall failure probability. His answer allows for this.
_____________________________________

About combining probabilities. Its easy to see the times when they simply get multiplied together - like say picking two winning numbers in a row.

Its different when you are trying to calculate the overall probability of a (usually unwanted) outcome given a probabilities of several things that might affect it.

Like say alike a power station. p1 the cooling system fails, p2 it gets struck by lightning, p3 a generator burns up, etc. Any one of these will stop the power from the plant.

How do you calculate the overall probability that the power stops 0<p<1 ?
_____________________________________
Yes, you have to take care when multiplying and working with probabilities.

First off, what does the multiplication operation mean with probabilities? The product p(A).p(B) represents the probability of both events occurring (the intersection), but if and only if they are independent. Independent means that one occurring will not affect the other.

In your example where single lottery balls are drawn then replaced, those single results are independent so you can freely multiply the probabilities. Drawing without replacement is slightly trickier, needs consideration of conditional probabilities. In your power plant example, bear in mind that p2 and p3 might be dependent! p2 directly implies there's a storm, and that might mean the pressure / temperature / humidity is skewed from normal, increasing the chance of p3. There's one more thing to point out (it's not a separate case from independence/dependence, rather an observation) - events can be mutually exclusive. Mutual exclusion is a form of dependence where one occurring means the other cannot; for example, in a draw without replacement, if your first card is the ace of spades, your second cannot be since those events are mutually exclusive.

The rule behind this is one of the three 'principles' of combinatorics called the Inclusion/Exclusion principle:

p(A or B or C or ...)
= p(A) + p(B) + p(C) + ...
- p(A and B) - p(A and C) - p(B and C) - ...
+ p(A and B and C) + ...
- (intersections of 4 events)
+ (intersections of 5 events) etc

Note that in probability we say union/intersection instead of or/and.
If you remove all the ellipsi from the general formula above, you'll get the one for three events:
p(A or B or C) = p(A) + p(B) + p(C) - p(A and B) - p(A and C) - p(B and C) + p(A and B and C)

The one for two events is:
p(A or B) = p(A) + p(B) - p(A and B)

The overall probability for the power plant stopping is just the inclusion/exclusion formula for 3 events. You need to obtain the 7 terms individually then use the formula to find the chance the union of the 3 events will happen. To illustrate, I'll make some arbitrary assumptions to cover a few cases:

1. p1 and p2 are independent - the cooling system failing has nothing to do with a lightning storm.
2. p1 implies p3 - the cooling system failing will ensure the generator burns (though it can also burn due to other things).
3. p2 and p3 overlap somewhat - the chance of the generator burning is high conditional on the power station being struck, and if the generator burned it was probably due to the station being struck.

The first 3 terms are probably known to you - the 3 probabilities on their own.
4th term; we know that p1, p2 are independent. we can simply multiply the probabilities to get this.
5th: Since p1 implies p3, the chance they both happen is just the chance of p1 on its own.
6th: This is where I guess you'd need some extra information. p(p2 and p3) can be rewritten as:
p(p2) + p(p3) - p(p2 or p3)
p(p2) - p(p2 and ~p3)
p(p3) - p(~p2 and p3)
Any of those might help in obtaining it.
7th: Since p1 implies p3, we can ignore p3 again. Since p1, p2 and independent, we can mlutiple p(p1) with p(p2) to get this term; in this case, p(p1 and p2 and p3) = p(p1 and p2) = p(p1).p(p2).

If two were mutually exclusive, their 'and' terms will be zero.
We hope this helps.

Lucky mkhonza
#4
Nov21-07, 06:41 AM
P: 14
Probability assessment

"What is the probability that the cooling circuit corresponding to the scenario arranged above fails?"

Let us assume component (1) Heat exchanger (probability = 10^(-4)/requirement), (2) Pump (probability = 10^(-3)/requirement) and energy to the pump (probability = 10^(-3)/requirement) are available to fulfil the function. How would you determine the probability above?

Since the three components above are available to fulfil the function then the system can only fail due to component 3 and 4. That is to say: If one of them fails: system fails (probability = (1-P3/4)*P3/4) or if both are not functioning then system fails (Probability = P3*P4).

Will this be correct?


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