Question

A complex electronic system is built with a certain number of backup components in its subsystems. One subsystem has eight identical components, each with a probability of 0.45 of failing in less than 1,000 hours. The sub system will operate if any four of the eight components are operating. Assume that the components operate independently. (Round your answers to four decimal places.)

Required:
Find the probability that the subsystem operates longer than 1000 hours.

Answer 1

0.7396 = 73.96% probability that the subsystem operates longer than 1000 hours.

Explanation:

For each component, there are only two possible outcomes. Either they fail in less than 1000 hours, or they do not. The components operate independently. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Eight components:

This means that
`n = 8`

Probability of 0.45 of failing in less than 1,000 hours.

So 1 - 0.45 = 0.55 probability of working for longer than 1000 hours, which means that
`p = 0.55`

Find the probability that the subsystem operates longer than 1000 hours.

We need at least four of the components operating. So

`P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 4) = C_(8,4).(0.55)^(4).(0.45)^(4) = 0.2627`

`P(X = 5) = C_(8,5).(0.55)^(5).(0.45)^(3) = 0.2568`

`P(X = 6) = C_(8,6).(0.55)^(6).(0.45)^(2) = 0.1569`

`P(X = 7) = C_(8,7).(0.55)^(7).(0.45)^(1) = 0.0548`

`P(X = 8) = C_(8,8).(0.55)^(8).(0.45)^(0) = 0.0084`

`P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2627 + 0.2568 + 0.1569 + 0.0548 + 0.0084 = 0.7396`

0.7396 = 73.96% probability that the subsystem operates longer than 1000 hours.