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

Answer:

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

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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 $\mu = \frac{1}{m}$ 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 $m = 5, \mu = \frac{1}{5} = 0.2$

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