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Exercise 3. The life of an electronic component has the exponential distri- bution with parameter = 0.0004 hrs-1. Assuming that the component is replaced as soon as it fails, find the probability that the number of components needed to accomplish a year of service is 0,1,2,3,4, and more than four.

Exercise 3. The life of an electronic component has the exponential distri- bution with parameter = 0.0004 hrs-1. Assuming that the component is replaced as soon as it fails, find the probability that the number of components needed to accomplish a year of service is 0,1,2,3,4, and more than four.

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Exercise 3
Theorem 4. If the time between indepenmdent events is exponentially dis- tributed with common parameter X, then the number of events in an interval of fixed length T is Poisson with parameter µ = XT. Proof Let Y, be the time until n arrivals, with density given by Equation (17), and let N be the number of arrivals in the interval [0,T). Then P(N > n) = P(Yn <T) (19) Think about it. And clearly P(N =n) = P(N > n) – P(N > n+ 1) (20) Hence P(N >n+ 1) = P(Yn+1 < T) = (21) dx = (22) n! Integrating by parts 7n-le-Ar dr = (23) n! (п — 1)! lo e-AT + P(Yn < T) (24) n! Equations (20) and (24) give the claimed expression for P(N = n). Exercise 3. The life of an electronic component has the exponential distri- bution with parameter = 0.0004 hrs-. Assuming that the component is replaced as soon as it fails, find the probability that the number of components needed to accomplish a year of service is 0,1,2,3,4, and more than four.
Transcribed Image Text:Theorem 4. If the time between indepenmdent events is exponentially dis- tributed with common parameter X, then the number of events in an interval of fixed length T is Poisson with parameter µ = XT. Proof Let Y, be the time until n arrivals, with density given by Equation (17), and let N be the number of arrivals in the interval [0,T). Then P(N > n) = P(Yn <T) (19) Think about it. And clearly P(N =n) = P(N > n) – P(N > n+ 1) (20) Hence P(N >n+ 1) = P(Yn+1 < T) = (21) dx = (22) n! Integrating by parts 7n-le-Ar dr = (23) n! (п — 1)! lo e-AT + P(Yn < T) (24) n! Equations (20) and (24) give the claimed expression for P(N = n). Exercise 3. The life of an electronic component has the exponential distri- bution with parameter = 0.0004 hrs-. Assuming that the component is replaced as soon as it fails, find the probability that the number of components needed to accomplish a year of service is 0,1,2,3,4, and more than four.
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