Question

When to replace a maintenance system. An article in the Journal of Quality of Maintenance Engineering (Vol. 19,2013) studied the problem of finding the optimal replacement policy for a maintenance system. Consider a system that is tested every 12 hours. The test will determine whether there are any flaws in the system. Assume that the probability of no flaw being detected is .85. If a flaw (failure) is detected, the system is repaired. Following the fifth failed test, the system is completely replaced. Now, let x represent the number of tests until the system needs to be replaced.

a. Give the probability distribution for x as a formula.

b. Find the probability that the system needs to be replaced after 8 total tests.

Short answer

a. The probability distribution for x as a formula is $\left(\begin{array}{l}x-1\\ x-5\end{array}\right){(1-0.85)}^5{\left(0.85\right)}^{x-5}$.

b. The probability that the system is replaced after 8 total tests is 0.0016.

Step-by-step solution

Calculating the Probability distribution for x with the formula

a.

Let x denote the random variable that the number of testes until the system need to be replaced.

The probability that no flaw is detected is$p=0.85$

And the probability that flaw is detected is

$\begin{array}{c}(1-p)=1-0.85\\ =0.15\end{array}$

The system is replaced if the 5^th flaw occurs on the last trail or on the xth trail.

Hence it is clear that the negative binomial distribution.

Let x be the discrete random variable with r number of failures, then probability mass function is,

localid="1661969502731" $\begin{array}{c}{\rm P}(X=x)=\left(\begin{array}{l}x-1\\ x-5\end{array}\right){(1-p)}^5{p}^{x-5};x=5,6,7\\ =\left(\begin{array}{l}x-1\\ x-5\end{array}\right){(1-0.85)}^5{\left(0.85\right)}^{x-5}\end{array}$

Calculating the Probability that the system need to be replaced after total 8 tests. 

b.

Consider,

$\begin{array}{c}{\rm P}(x=8)=\left(\begin{array}{l}8-1\\ 8-5\end{array}\right){(1-0.85)}^5{\left(0.85\right)}^{8-5}\\ =\left(\begin{array}{l}8-1\\ 8-5\end{array}\right){\left(0.15\right)}^5{\left(0.85\right)}^3\\ =\left((\begin{array}{l}7\\ 3\end{array}\right){\left(0.15\right)}^5{\left(0.85\right)}^3\end{array}$

$\begin{array}{c}{\rm P}(x=8)=35\times 0.0000759375\times 0.614125\\ =0.0016\end{array}$

Hence the probability that the system is replaced after 8 total tests is 0.0016.