Question

Suppose that a certain system contains three components that function independently of each other and are connected in series, as defined in Exercise 5 of Sec. 3.7, so that the system fails as soon as one of the components fails. Suppose that the length of life of the first component, measured in hours, has the exponential distribution with parameter\(\beta = 0.001\), the length of life of the second component has the exponential distribution with parameter\(\beta = 0.003\), and the length of life of the third component has the exponential distribution with parameter\(\beta = 0.006\).Determine the probability that the system will not fail before 100 hours.

Short answer

Hence, probability that the system will not fail before 100 hours is 0.3679.

Step-by-step solution

Given information

A system contains three components that function independently of each other. Length of all the components has the exponential distribution.

Computing the probability

Since, length of the first component follows exponential distribution with parameter \(\beta = 0.001\), length of the second component follows exponential distribution with parameter \(\beta = 0.003\) and length of the third component follows exponential distribution with parameter \(\beta = 0.006\).

Let X be the length of life of the system.

Then,

X has the exponential distribution with parameter \(\left( {0.001 + 0.003 + 0.006} \right) = 0.01\)

Since, by the definition of exponential distribution,

\(P\left( {X > x} \right) = {e^{ - \beta x}}\)

Therefore,

\(\begin{aligned}{c}{\rm P}\left( {X > 100} \right)& = {e^{ - 100(0.01)}}\\ &= {e^{ - 1}}\\ &= 0.3679\end{aligned}\)

Hence, probability that the system will not fail before 100 hours is 0.3679.