An engineering system consisting of n components is said to be a k-out-of-n syst
ID: 3365806 • Letter: A
Question
An engineering system consisting of n components is said to be a k-out-of-n system, that is, the system functions if and only if at least k of the n components functions well. Suppose all a) If each component independently works with the same probability 0.8, compute the b) Assume each component independently works with different probability. The probability the components function independently of each other. probability that a 3-out-of-5 system functions. +1 i = 1,2, ,5, compute the +2 probability that a 4-out-of-5 system functions. c) For a 2-out-of-4 system, if each component independently works with the same probability 0.9, find out the probability that at least 3 components could work given that the 2-out-of-4 system works.Explanation / Answer
(a) Each component works with probability 0.8 and fails with probability 0.2. This is a binomial distribution.
Probability that a 3-out-of-5 system functions = Probability that exactly 3 components work + Probability that exactly 4 components work + Probability that exactly 5 components work
= C [5,3] x (0.8)3 x (0.2)2 + C [5,4] x (0.8)4 x (0.2)1 + C [5,5] x (0.8)5 x (0.2)0 = 10 x 0.512 x 0.05 + 5 x 0.4096 x 0.2 + 1 x 0.32768 x 1 = 0.2048 + 0.4096 + 0.32768 = 0.94208
(b) Probability that a 4-out-of-5 system functions = Probability that 4 components function + Probability that all 5 components function
Probability that 4 components function = P1 x P2 x P3 x P4 x (1-P5) + P1 x P2 x P3 x (1-P4 ) x P5 + P1 x P2 x (1-P3 ) x P4 x P5 + P1 x (1-P2 ) x P3 x P4 x P5 + (1-P1 ) x P2 x P3 x P4 x P5 = (2/3) x (3/4) x (4/5) x (5/6) x (1-6/7) + (2/3) x (3/4) x (4/5) x (1-5/6) x (6/7) + (2/3) x (3/4) x (1-4/5) x (5/6) x (6/7) + (2/3) x (1-3/4) x (4/5) x (5/6) x (6/7) + (1-2/3) x (3/4) x (4/5) x (5/6) x (6/7) = 2 X (1/6) x (1/7) + 2 x (1/5) x (1/7) + 2 x (1/4) x (1/7) + (2/3) x (1/7) + (1/3) x (3/7) = 1/21 + 2/35 + 1/14 + 2/21 + 1/7 = 0.41428
Probability that all 5 components function = P1 x P2 x P3 x P4 x P5 = 2/3 x 3/4 x 4/5 x 5/6 x 6/7 = 2/7 = 0.28571
Probability that a 4-out-of-5 system functions = 0.7
(c) Probability that at least 3 components work = Probability that 3 components work + Probability that all 4 components work = C [4,3] x (0.9)3 x (0.1) + C [4,4] x (0.9)4 = 4 x 0.729 x 0.1 + 1 x 0.6561 = 0.2916 + 0.6561 = 0.9477
Probability that a 2-out-of-4 system works = Probability that 2 components work + workProbability that 3 components work + Probability that all 4 components work = C [4,2] x (0.9)2 x (0.1)2 = 6 x 0.81 x 0.01 + 0.9477 = 0.0486 = 0.9963
So, the Probability that at least 3 components work, given that a 2-out-of-4 system works = 0.9477/0.9963 = 0.9512