Basic probability theory Professor Jrn Vatn 1 Event
Basic probability theory Professor Jørn Vatn 1
Event n Probability relates to events n 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. : n A = {operator error next year} n B = {component failure next year} n An event may occur, or not. We do not know that in advance prior to the experiment or a situation in the “real life”. 2
Probability n When events are defined, the probability that the event will occur is of interest n Probability is denoted by Pr(·), i. e. n Pr(A) = Probability that A (will) occur n The numeric value of Pr(A) may be found by: n Studying the sample space / symmetric considerations n Analysing collected data n Look up the value in data hand books n “Expert judgement” n Laws of probability calculus/Monte Carlo simulation 3
Sample space n The sample space defines all possible events n As an example let A = {It is Sunday}, B = {It is Monday}, . . , G = {It is Saturday}. The sample space is then given by n S = {A, B, C, D, E, F, G} n So-called Venn diagrams are useful when we want to analyze subset of the sample space S. 4
Venn diagram n A rectangle represents the sample space, and closed curves such as a circle are used to represent subsets of the sample space 5
Union n The union of two events A and B: n A B denotes the occurrence of A or B or (A and B) n Example n A = {prime numbers 6) n B = {odd numbers 6} n A B = {1, 2, 3, 5} 6
Intersection n The intersection of two events A and B: n A B denotes the occurrence of both A and B n Example n A = {prime numbers 6) n B = {odd numbers 6} n A B = {3, 5} 7
Disjoint events n A and B are said to be disjoint if they can not occur simultaneously, i. e. A B = Ø = the empty set 8
Complementary events n The complement of an event A is all events in the sample space S except for A. n The complement of an event A is denoted by AC n Example n A = {even numbers) n AC = {odd numbers} 9
Probability n Probability is a set function Pr() which maps events A 1, A 2, . . . in the sample space S, to real numbers n 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 A 2 S 0 P(A 1) P(A 2) 1 10
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 11
Conditional probability n In some situations the probability of A will change if we get information about a related event, say B n We then introduce conditional probabilities, and write: n Pr(A|B) = the conditional probability that A will occur given that B has occurred n Example: Probability of pulling ace of spade is 1/52, but if we have seen a “black” card, the conditional probability is 1/26 12
Independent events n A and B are said to be independent if information about whether B has occurred does not influence the probability that A will occur n Pr(A|B) = Pr(A) n 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) 13
Basic rules for probability calculus n Pr(A B) = Pr(A) + Pr(B) - Pr(A B) n Pr(A B) = Pr(A) Pr(B) if A and B are independent Pr(AC) = Pr(A does not occur) = 1 - Pr(A) n Pr(A|B) = Pr(A B) / Pr(B) 14
Example n Let A = {It is Sunday} n B = {It is between 6 and 8 pm) n A and B are independent but not disjoint n We will find Pr(A B) and Pr(A B) n Pr(A B) = Pr(A) Pr(B) = n Pr(A B) = Pr(A)+ Pr(B) - Pr(A B) = n Pr(A|B) = 15
Example n Assume we have two redundant shut-down valves, ESDV and PSDV that could be used in an emergency situation n Pr(ESDV-failure)=0. 01 n Pr(PSDV-failure)=0. 005 n Assuming independent failures give a total failure probability of n 0. 01 0. 005 = 5 10 -5 16
Division of the sample space n A 1, A 2, …, Ar is said to be a division of the sample space if the union of all Ai’s covers the entire sample space, i. e. A 1 A 2 … Ar = S and the Ais are pair wise disjoint, Ai Aj = Ø for i j 17
The law of total probability n Let A 1, A 2, …, Ar represent a division of the sample space S, and let B be an arbitrary event in S, then 18
Example n A special component type is ordered from two suppliers A 1 and A 2 n Experience has shown that n components from supplier A 1 has a defect probability of 1% n components from supplier A 2 has a defect probability of 2% n In average 70% of the components are provided by supplier A 1 n 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 n A component is now fetched from the stock, and we will calculate the defect probability, Pr(B) 19
Exercise n Successful evacuation depends on the available evacuation time, n A 1 = short evacuation time Pr(A 1) = 1% n A 2 = medium evacuation time Pr(A 2) = 20% n A 3 = long evacuation time Pr(A 3) = 79% n The probability of successful evacuation (B) is given by: n Pr(B| A 1) = 50% n Pr(B| A 2) = 75% n Pr(B| A 3) = 95% n Find Pr(B) by the law of total probability 20
Random quantities n A random quantity (stochastic variable), is a quantity for which we do not know the value it will take, but n We could state statistical properties of the quantity or make probability statement about it n Whereas an event may occur, or not occur (B&W), a random quantity is related to a magnitude, it may take different values n We use probabilities to describe the likelihood of the different values the random quantity can take n Cumulative distribution function (S-curve) n Probability density function (histogram) 21
Examples of random quantities n n n 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 22
Cumulative distribution function (CDF) FX(x) = Pr(X x) 23
Exercise n Let X be the life time of a component 2 -(0. 01 x) n Use Excel to find Pr(X 150) when FX(x) = 1 - e 24
Probability density function (PDF) 25
PDF probabilities 26
Expectation n 27
Variance n 28
Standard deviation 29
Parameters describing random quantities n n n Percentiles, i. e. P 1, P 10, P 50, P 99 Most likely value (M) Expected (mean) value ( ) Standard deviation ( ) Variance (Var = 2) 30
Expectation and variance for a sum n 31
Life times n In reliability theory we work with life times n The life time, or time to failure, is the time it takes from a component is installed, until it fails for the first time n Life times are non-negative random quantities n For life times we introduce the following concepts n R(x) = Pr(X > x) = 1 - FX(x) n MTTF = Mean Time To Failure = E(X) 32
Statistical view of life times 33
Distribution classes n Life times are often associated with various distribution classes, e. g. in reliability analysis we often apply the following distribution classes n The exponential distribution n The Weibull distribution n The gamma distribution n The normal distribution 34
The exponential distribution n The exponential distribution is a very simple distribution which could be used if no aging affects the component under consideration n Often external or internal shocks dominates the failure causes if the exponential distribution is used n For the exponential distribution we have n f. X(x) = e- x n FX(x) = 1 -e- x n R(x) = e- x n E(X) =1/ n Var(X) = 1/ 2 n is a parameter in the distribution (the failure rate) 35
Example n We will obtain the probability that X is greater than it’s expected value. We then have: n Pr(X > E(X)) = R(E(X)) = e- E(X ) = e -1 0. 37 n i. e. , most likely it will not survive the expected life time 36
Example n Assume the life time, X, of a component is exponentially distributed with parameter = 0. 01 n 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 = Pr(X > 200) n Thus, an old component is stochastically as good as a new component 37
For the Weibull distribution we have -( x) e n R(x) = n is a shape parameter, > 1 means aging n MTTF = n Var(X) = n where ( ) is the gamma function n The Gamma function is found in Excel by =EXP(GAMMALN(x)) 38
Reparameterization of the Weibull n The Weibull distribution has two parameters: n = shape or aging parameter n = scale parameter n The relation between and MTTF is n MTTF = n In many situations it is easier to work with and MTTF, rather than and 39
Example n 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 n Pr(X > 400 |X > 200) = Pr(X > 400 X > 200)/Pr(X > 200) = Pr(X > 400)/Pr(X > 200) = 2 2 2 R(400)/R(200) = e-( 400) / e-( 200) e-( 350) < Pr(X > 200) n Thus, an old component is not as good as a new one 40
The hazard rate, z(t) n The hazard rate is the precise term for the so-called bathtub curve, also denoted failure rate funciton: n z(t) = f(t)/R(t) n z(t) t Probability of failure in a small time interval ( t ) given that the unit has survived up to t. 41
Example of hazard rates n Exponential distribution n z(t) = = constant n Weibull distribution n z(t) = ( )( t) -1 t -1 = increasing in time t for > 1 n Preventive maintenance is often based on the idea of ”taking away” the right hand side of the hazard rate curve 42
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