Question

Consider a series system composed of 4 separate components where each component has a 30% chance of failing. Assume each component functions independent of each other and the system will fail if at least one of the components fails. If you have 7 of these systems, what is the probability that exactly 3 of them would function? Clearly state the events of interest and random variable of interest using the context of the problem. Hint: Find the probabi

Answer 1

16.15% probability that exactly 3 of them would function

Explanation:

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.

Probability of each system working:

4 components, which means that
`n = 4`

Each has a 30% probability of failing, so
`p = 1 - 0.3 = 0.7`

For the system to work, all 4 components have to work. This is P(X = 4).

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

`P(X = 4) = C_(4,4).(0.7)^(4).(0.3)^(0) = 0.2401`

0.2401 probability of a system working.

If you have 7 of these systems, what is the probability that exactly 3 of them would function?

Now 7 systems, so
`n = 7`

0.2401 probability of a system working.

We have to find P(X = 3).

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

`P(X = 3) = C_(7,3).(0.2401)^(3).(0.7599)^(4) = 0.1615`

16.15% probability that exactly 3 of them would function