1. 1 Basic probability theory Professor Jørn Vatn

2. 2 Event Probability relates to events Let as an example A be the event that there is an operator error in a control room next year, and B be the event that there is a specific component failure next year i.e.: A = {operator error next year} B = {component failure next year} An event may occur, or not. We do not know that in advance prior to the experiment or a situation in the “real life”.

3. 3 Probability When events are defined, the probability that the event will occur is of interest Probability is denoted by Pr(·), i.e. Pr(A) = Probability that A (will) occur The numeric value of Pr(A) may be found by: Studying the sample space / symmetric considerations Analysing collected data Look up the value in data hand books “Expert judgement” Laws of probability calculus/Monte Carlo simulation

4. 4 Sample space The sample space defines all possible events As an example let A = {It is Sunday}, B = {It is Monday},.., G = {It is Saturday}. The sample space is then given by S = {A,B,C,D,E,F,G} So-called Venn diagrams are useful when we want to analyze subset of the sample space S.

5. 5 Venn diagram A rectangle represents the sample space, and closed curves such as a circle are used to represent subsets of the sample space

6. 6 Union The union of two events A and B: A  B denotes the occurrence of A or B or (A and B) Example A = {prime numbers  6) B = {odd numbers  6} A  B = {1,2,3,5}

7. 7 Intersection The intersection of two events A and B: A  B denotes the occurrence of both A and B Example A = {prime numbers  6) B = {odd numbers  6} A  B = {3,5}

8. 8 Disjoint events A and B are said to be disjoint if they can not occur simultaneously, i.e. A  B = Ø = the empty set

9. 9 Complementary events The complement of an event A is all events in the sample space S except for A. The complement of an event A is denoted by A C Example A = {even numbers) A C = {odd numbers}

10. 10 Probability Probability is a set function Pr() which maps events A 1, A 2,... in the sample space S, to real numbers The function Pr(  ) can only take values in the interval from 0 to 1, i.e. probabilities are greater or equal than 0, and less or equal than 1 S 01 A 1 A 2 P(A 1 ) 2 )

11. 11 Kolmogorov basic axioms 1.0  Pr(A) 2.Pr(S) = 1 3.If A 1, A 2,... is a sequence of disjoint events we shall then have Pr(A 1  A 2 ...) = Pr(A 1 ) + Pr(A 2 ) +... Everything is based on these axioms in probability calculus

12. 12 Conditional probability In some situations the probability of A will change if we get information about a related event, say B We then introduce conditional probabilities, and write: Pr(A|B) = the conditional probability that A will occur given that B has occurred Example: Probability of pulling ace of spade is 1/52, but if we have seen a “black” card, the conditional probability is 1/26

13. 13 Independent events A and B are said to be independent if information about whether B has occurred does not influence the probability that A will occur Pr(A|B) = Pr(A) Example: We are both pulling a card and tossing a dice in a composed experiment. The probability of pulling ace of spade (A) is independent of the event getting a six (B)

14. 14 Basic rules for probability calculus Pr(A  B) = Pr(A) + Pr(B) - Pr(A  B) Pr(A  B) = Pr(A)  Pr(B) if A and B are independent Pr(A C ) = Pr(A does not occur) = 1 - Pr(A) Pr(A|B) = Pr(A  B) / Pr(B)

15. 15 Example LetA = {It is Sunday} B = {It is between 6 and 8 pm) A and B are independent but not disjoint We will find Pr(A  B) and Pr(A  B) Pr(A  B) = Pr(A)  Pr(B) = Pr(A  B) = Pr(A)+ Pr(B) - Pr(A  B) = Pr(A|B) =

16. 16 Example Assume we have two redundant shut-down valves, ESDV and PSDV that could be used in an emergency situation Pr(ESDV-failure)=0.01 Pr(PSDV-failure)=0.005 Assuming independent failures give a total failure probability of 0.01  0.005 = 5  10 -5

17. 17 Division of the sample space A 1,A 2,…,A r is said to be a division of the sample space if the union of all A i ’s covers the entire sample space, i.e. A 1  A 2  …  A r = S and the A i s are pair wise disjoint, A i  A j = Ø for i  j

18. 18 The law of total probability Let A 1,A 2,…,A r represent a division of the sample space S, and let B be an arbitrary event in S, then

19. 19 Example A special component type is ordered from two suppliers A 1 and A 2 Experience has shown that components from supplier A 1 has a defect probability of 1% components from supplier A 2 has a defect probability of 2% In average 70% of the components are provided by supplier A 1 Assume that all components are put on a common stock, and we are not able to trace the supplier for a component in the stock A component is now fetched from the stock, and we will calculate the defect probability, Pr(B)

20. 20 Exercise Successful evacuation depends on the available evacuation time, A 1 = short evacuation time  Pr(A 1 ) = 1% A 2 = medium evacuation time  Pr(A 2 ) = 20% A 3 = long evacuation time  Pr(A 3 ) = 79% The probability of successful evacuation (B) is given by: Pr(B| A 1 ) = 50% Pr(B| A 2 ) = 75% Pr(B| A 3 ) = 95% Find Pr(B) by the law of total probability

21. 21 Random quantities A random quantity (stochastic variable), is a quantity for which we do not know the value it will take, but We could state statistical properties of the quantity or make probability statement about it Whereas an event may occur, or not occur (B&W), a random quantity is related to a magnitude, it may take different values We use probabilities to describe the likelihood of the different values the random quantity can take Cumulative distribution function (S-curve) Probability density function (histogram)

22. 22 Examples of random quantities X = Life time of a component (continuous) R = Repair time after a failure (continuous) Z = Number of failures in a period of one year (discrete) M = Number of derailments next year N = Number of delayed trains next month W = Maintenance cost next year

23. 23 Cumulative distribution function (CDF) F X (x) = Pr(X  x)

24. 24 Exercise Let X be the life time of a component Use Excel to find Pr(X  150) when F X (x) = 1 - e -(0.01x) 2

25. 25 Probability density function (PDF)

26. 26 PDF  probabilities

27. 27 Expectation

28. 28 Variance

29. 29 Standard deviation

30. 30 Parameters describing random quantities Percentiles, i.e. P1,P10,P50,P90,P99 Most likely value (M) Expected (mean) value (  ) Standard deviation (  ) Variance (Var =  2 )

31. 31 Expectation and variance for a sum

32. 32 Life times In reliability theory we work with life times The life time, or time to failure, is the time it takes from a component is installed, until it fails for the first time Life times are non-negative random quantities For life times we introduce the following concepts R(x) = Pr(X > x) = 1- F X (x) MTTF = Mean Time To Failure = E(X)

33. 33 Statistical view of life times

34. 34 Distribution classes Life times are often associated with various distribution classes, e.g. in reliability analysis we often apply the following distribution classes The exponential distribution The Weibull distribution The gamma distribution The normal distribution

35. 35 The exponential distribution The exponential distribution is a very simple distribution which could be used if no aging affects the component under consideration Often external or internal shocks dominates the failure causes if the exponential distribution is used For the exponential distribution we have f X (x) = e - x F X (x) = 1-e - x R(x) = e - x E(X) =1/ Var(X) = 1/ 2 is a parameter in the distribution (the failure rate)

36. 36 Example We will obtain the probability that X is greater than it’s expected value. We then have: Pr(X > E(X)) = R(E(X)) = e - E(X ) = e -1  0.37 i.e., most likely it will not survive the expected life time

37. 37 Example Assume the life time, X, of a component is exponentially distributed with parameter = 0.01 We will find the probability that the component that has survived 200 hours, will survive another 200 hours Pr(X > 400 |X > 200) = Pr(X > 400  X > 200)/Pr(X > 200) = Pr(X > 400)/Pr(X > 200) = R(400)/R(200) = e -  400 / e -  200 = e -  200 = Pr(X > 200) Thus, an old component is stochastically as good as a new component

38. 38 For the Weibull distribution we have R(x) = e -( x)   is a shape parameter,  > 1 means aging MTTF = Var( X ) = where  (  ) is the gamma function The Gamma function is found in Excel by =EXP(GAMMALN(x))

39. 39 Reparameterization of the Weibull The Weibull distribution has two parameters:  = shape or aging parameter  = scale parameter The relation between and MTTF is MTTF = In many situations it is easier to work with  and MTTF, rather than  and

40. 40 Example We will find the probability that a component that has survived 200 hours, will survive another 200 hours given that the life time is Weibull distributed with parameter  = 2 and = 0.01 Pr(X > 400 |X > 200) = Pr(X > 400  X > 200)/Pr(X > 200) = Pr(X > 400)/Pr(X > 200) = R(400)/R(200) = e -(  400) 2 / e -(  200) 2  e -(  350) 2 200) Thus, an old component is not as good as a new one

41. 41 The hazard rate, z(t) The hazard rate is the precise term for the so-called bathtub curve, also denoted failure rate funciton: z(t) = f(t)/R(t) z(t)  t  Probability of failure in a small time interval (  t ) given that the unit has survived up to t.

42. 42 Example of hazard rates Exponential distribution z(t) = = constant Weibull distribution z(t) = (  )( t)  -1   t  -1 = increasing in time t for  > 1 Preventive maintenance is often based on the idea of ”taking away” the right hand side of the hazard rate curve