rpra3 - Engineering Risk Benefit Analysis 1.155, 2.943,...

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RPRA 3. Probability Distributions in RPRA 1 Engineering Risk Benefit Analysis 1.155, 2.943, 3.577, 6.938, 10.816, 13.621, 16.862, 22.82, ESD.72, ESD.721 RPRA 3. Probability Distributions in RPRA George E. Apostolakis Massachusetts Institute of Technology Spring 2007
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RPRA 3. Probability Distributions in RPRA 2 Overview We need models for: The probability that a component will start (fail) on demand. The probability that a component will run for a period of time given a successful start. The impact of repair on these probabilities. The frequency of initiating events.
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RPRA 3. Probability Distributions in RPRA 3 Failure to start P[failure to start on demand] q unavailability P[successful start on demand] p availability Requirement: p + q = 1
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RPRA 3. Probability Distributions in RPRA 4 The Binomial Distribution (1) Start with an “experiment” that can have only two outcomes: “success” and “failure” or {0, 1} with probabilities p and q, respectively. Consider N "trials," i.e., repetitions of this experiment with constant q. These are called Bernoulli trials . Define a new DRV: X = number of 1's in N trials Sample space of X: {0,1,2,. ..,N} What is the probability that there will be k 1’s (failures) in N trials?
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RPRA 3. Probability Distributions in RPRA 5 The Binomial Distribution (2) This is the probability mass function of the Binomial Distribution. It is the probability of exactly k failures in N demands. The binomial coefficient is : k N k ) q 1 ( k N ] k X Pr[ q = = )! k N ( ! k ! N k N For the coin : Assume 3 trials. We are interested in 1 failure. There are 3 such sequences: fss, sfs, ssf (mutually exclusive). The probability of each is qp 2 . If order is unimportant, the probability of 1 failure in 3 trials is 3qp 2 .
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RPRA 3. Probability Distributions in RPRA 6 The Binomial Distribution (3) Mean number of failures: qN Variance: q(1-q)N Normalization: P[at most m failures] = 1 ) q 1 ( k N N 0 k k N k q = = CDF ) m ( F ) q 1 ( k N m 0 k k N k q =
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RPRA 3. Probability Distributions in RPRA 7 Example: 2-out-of-3 system We found in slide 16 of RPRA 1 that the structure function is (using min cut sets): X T = ( X A X B + X B X C + X C X A ) - 2 X A X B X C The failure probability is P(failure) = P ( X T =1 ) = 3q 2 –2q 3 Using the binomial distribution: Pr(system failure) = P[2 fail] + P[3 fail] = 3q 2 (1-q) + q 3 = 3q 2 3 Notes : 1. We have assumed nominally identical and independent components in both cases. 2. If the components are not nominally identical or independent, the structure function approach still works, but the binomial distribution is not applicable. Why?
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RPRA 3. Probability Distributions in RPRA 8 The Poisson Distribution Used typically to model the occurrence of initiating events.
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This note was uploaded on 11/08/2011 for the course AERO 16.851 taught by Professor Ldavidmiller during the Fall '03 term at MIT.

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rpra3 - Engineering Risk Benefit Analysis 1.155, 2.943,...

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