An engineer designs a circuit, seen below. In this system, if the first component fails, the functionality moves immediately to the second component and so on. The system only fails when the fifth component fails. Each component has a lifetime that is exponentially distributed with λ = 0.01 and components fail independently of one another. Define A, to be the length of time component i lasts. Let Y = the time at which the new system fails. Given N m • A-Exponential (0.01) • Y=A1+A2+A3+A4+A5 1. Find the probability that this new system will last fewer than 50 hours.

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An engineer designs a circuit, seen below. In this system, if the first component fails, the
functionality moves immediately to the second component and so on. The system only fails
when the fifth component fails. Each component has a lifetime that is exponentially distributed
with λ = 0.01 and components fail independently of one another. Define A, to be the length of
time component i lasts. Let Y = the time at which the new system fails.
Given
2
لنا
5
A-Exponential (0.01)
Y=A1+A2+A3+A4+A5
1. Find the probability that this new system will last fewer than 50 hours.
Transcribed Image Text:An engineer designs a circuit, seen below. In this system, if the first component fails, the functionality moves immediately to the second component and so on. The system only fails when the fifth component fails. Each component has a lifetime that is exponentially distributed with λ = 0.01 and components fail independently of one another. Define A, to be the length of time component i lasts. Let Y = the time at which the new system fails. Given 2 لنا 5 A-Exponential (0.01) Y=A1+A2+A3+A4+A5 1. Find the probability that this new system will last fewer than 50 hours.
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