Question

An organization purchases 10 computers from a manufacturer with lifetimes modeled as exponential variables (mean of 5 years). What is the probability that at least 6 computers are functioning beyond 6 years

Answer 1

0.0482 = 4.82% probability that at least 6 computers are functioning beyond 6 years

Explanation:

For each computer, there are only two possible outcomes. Either it works beyond 6 years, or it does not. The probability of a computer working beyong 6 years is independent of any other computes. This means that the binomial probability distribution is used to solve this question.

As the lifetime of a computer is an exponential variable, the probability of a computer working beyond 6 years is found using the exponential distribution.

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.

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

`f(x) = \mu e^(-\mu x)`

In which
`\mu = (1)/(m)` is the decay parameter.

The probability that x is lower or equal to a is given by:

`P(X \leq x) = \int\limits^a_0 {f(x)} \, dx`

Which has the following solution:

`P(X \leq x) = 1 - e^(-\mu x)`

The probability of finding a value higher than x is:

`P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^(-\mu x)) = e^(-\mu x)`

Probability of a computer working beyong 6 years:

Mean of 5 years means that
`m = 5, \mu = (1)/(5) = 0.2`

The probability is:

`P(X > 6) = e^(-0.2*6) = 0.3012`

What is the probability that at least 6 computers are functioning beyond 6 years?

10 computers means that
`n = 10`

0.3012 probability of a computer working beyond 6 years means that
`p = 0.3012`

This probability is:

`P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)`

In which

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

`P(X = 6) = C_(10,6).(0.3012)^(6).(0.6988)^(4) = 0.0374`

`P(X = 7) = C_(10,7).(0.3012)^(7).(0.6988)^(3) = 0.0092`

`P(X = 8) = C_(10,8).(0.3012)^(8).(0.6988)^(2) = 0.0015`

`P(X = 9) = C_(10,9).(0.3012)^(9).(0.6988)^(1) = 0.0001`

`P(X = 10) = C_(10,10).(0.3012)^(10).(0.6988)^(0) \approx 0`

`P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0374 + 0.0092 + 0.0015 + 0.0001 + 0 = 0.0482`

0.0482 = 4.82% probability that at least 6 computers are functioning beyond 6 years