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amilapsn

  • one year ago

An explosion at a construction site could have occurred as the result of static electricity, malfunctioning of equipment, carelessness or sabotage. Inter-views with construction engineers analysing the risks involved led to the estimates that such an explosion would occur with probability 0.25 as a result of static electricity, 0.20 as a result of malfunctioning of equipment, 0.40 as a result of carelessness and 0.75 as a result of sabotage. It is also felt the prior probabilities of the four causes of the explosion are 0.20, 0.40, 0.25 and 0.15. Based on all the information....

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  1. amilapsn
    • one year ago
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    (a) What is the most likely cause of the explosion? (b) What is the least likely cause for the explosion?

  2. amilapsn
    • one year ago
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    Am I correct?: "Such an explosion would occur with probability 0.25 as a result of static electricity " \(\equiv\) P(explosion/static electricity)=0.25

  3. amilapsn
    • one year ago
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    Or is it the other way round?

  4. kropot72
    • one year ago
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    Surely "0.75 as a result of sabotage" should have been written as "0.15 as a result of sabotage". Please confirm.

  5. amilapsn
    • one year ago
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    I don't get it @kropot72

  6. amilapsn
    • one year ago
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    Are you telling the given probabilities are the same?

  7. kropot72
    • one year ago
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    The sum of the given prior probabilities is 1. The sum of the posterior probabilities must also be 1 (by the basic rule of probabilities). The most likely error in the posterior probabilities is "...0.75 as a result of sabotage". If this is written as "...0.15 as a result of sabotage", the sum of the posterior probabilities would be 1.

  8. kropot72
    • one year ago
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    It is quite possible for the prior and the posterior probabilities of sabotage to both be 0.15.

  9. amilapsn
    • one year ago
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    But in the question it's 0.75

  10. amilapsn
    • one year ago
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    Am I correct?: "Such an explosion would occur with probability 0.25 as a result of static electricity " \(\equiv\) P(explosion/static electricity)=0.25 Or is it the other way round?

  11. kropot72
    • one year ago
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    The solution is found from Bates Theorem: \[\large P(A|B)=\frac{P(A) P(B|A)}{P(B)}\] for proposition A, and evidence B where P(A) is the prior, the initial degree of belief in A P(A|B) is the posterior, the degree of belief having accounted for B and the quotient P(B | A)/P(B) represents the support B provides for A. So if we consider static electricity as a possible cause, the support the evidence provides for the prior is given by 0.25/0.2 = 1.25. The support for each of the other three possible causes is found in a similar way.

  12. kropot72
    • one year ago
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    "But in the question it's 0.75" Based on the reasons in my previous posting, you need to query this part of the question with your lecturer.

  13. amilapsn
    • one year ago
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    ok....

  14. kropot72
    • one year ago
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    "The solution is found from Bates Theorem:" should obviously read "The solution is found from Bayes Theorem:"

  15. amilapsn
    • one year ago
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    Is this the answer: A: Static Electricity B: Malfunctioning of equipment C: Carelessness D: Sabotage E: Explosion P(E|A)=0.25

  16. amilapsn
    • one year ago
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    P(E|B)=0.20 P(E|C)=0.40 P(E|D)=0.15 (As you suggested)

  17. amilapsn
    • one year ago
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    P(A)=0.20 P(B)=0.40 P(C)=0.25 P(D)=0.15

  18. amilapsn
    • one year ago
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    We have to find P(A|E), P(B|E), P(C|E) and P(D|E) right?

  19. amilapsn
    • one year ago
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    P(A|E)=P(A and E)/P(E), ok?

  20. amilapsn
    • one year ago
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    But to find that we need to find P(E)...

  21. amilapsn
    • one year ago
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    Hey I think P(E|D) can be equal to 0.75...

  22. amilapsn
    • one year ago
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    Because, P(E|A)+P(E|B)+P(E|C)+P(E|D) should not be 1

  23. amilapsn
    • one year ago
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    *should not necessarily be

  24. amilapsn
    • one year ago
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    The given probabilities are not posterior probabilities...

  25. kropot72
    • one year ago
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    The first part of the question is: (a) What is the most likely cause of the explosion? So you need to find what support the evidence has on each of the prior probabilities. For static electricity the support is 1.25. For malfunction the support is 0.5. For carelessness the support is 1.6. For sabotage the support is 1 (assuming the posterior is 0.15 and not 0.75). Before taking the evidence, the leading possibilities were malfunction and carelessness, in that order. However in the light of the evidence carelessness is now the most likely cause. (b) In the light of the evidence sabotage is the least likely cause.

  26. kropot72
    • one year ago
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    "The given probabilities are not posterior probabilities..." The last set of probabilities are given as prior probabilities. How was it concluded that the evidence -based probabilities are not posterior probabilities?

  27. amilapsn
    • one year ago
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    Because the phrase, "...an explosion would occur with probability 0.25 as a result of static electricity..." suggests P(Explosion|Static Electricity)=0.25 \(\sf \underline{not}\) P(Static Electricity|Explosion)

  28. amilapsn
    • one year ago
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    i. e. it suggest the probability of an explosion if there is static electricity.

  29. kropot72
    • one year ago
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    Sorry to say I need to log out now. You need to study the application of Bayes' Theorem to this kind of question. The first set of probabilities in the question are definitely posterior probabilities, based on evidence. You could find the information here helpful: https://en.wikipedia.org/wiki/Bayes'_theorem

  30. amilapsn
    • one year ago
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    np.

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