Question

9. A large electronic office product contains 2000 electronic components. Assume that the probability that each component operates without failure during the useful life of the product is 0.995, and assume that the components fail independently. Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

Answer 1

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.

Explanation:

For each component, there are only two possible outcomes. Either it works correctly, or it does not. The probability of a component falling is independent from other components. 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.

In this problem we have that:

`n = 2000, p = 1-0.995 = 0.005`

Approximate the probability that five or more of the original 2000 components fail during the useful life of the product.

We know that either less than five compoenents fail, or at least five do. The sum of the probabilities of these events is decimal 1. So

`P(X < 5) + P(X \geq 5) = 1`

We want
`P(X \geq 5)`

So

`P(X \geq 5) = 1 - P(X < 5)`

In which

`P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)`

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

`P(X = 0) = C_(2000,0).(0.005)^(0).(0.995)^(2000) = 0.000044`

`P(X = 1) = C_(2000,1).(0.005)^(1).(0.995)^(1999) = 0.000445`

`P(X = 2) = C_(2000,2).(0.005)^(2).(0.995)^(1998) = 0.002235`

`P(X = 3) = C_(2000,3).(0.005)^(3).(0.995)^(1997) = 0.007480`

`P(X = 4) = C_(2000,4).(0.005)^(4).(0.995)^(1996) = 0.018765`

`P(X < 5) = P(X = 0) + `P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.000044 + 0.000445 + 0.002235 + 0.007480 + 0.018765 = 0.0290`

`P(X \geq 5) = 1 - P(X < 5) = 1 - 0.0290 = 0.9710`

97.10% probability that five or more of the original 2000 components fail during the useful life of the product.