Question

It is known that screws produced by a certain company will be defective with probability 0.01, independently of each other. The company sells the screws in packages of 10 and offers a money-back guarantee that at most 1 of 10 screws are defective. That is, the package would have to be replaced if greater or equal to 2 screws are defective. What proportion of packages sold must the company replace

Answer 1

The company must replace 0.0042 = 0.42% of packages.

Explanation:

For each screw, there are only two possible outcomes. Either it is defective, or it is not. The probability of an screw being defective is independent of any other screw. This means that the binomial probability distribution is used 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.

It is known that screws produced by a certain company will be defective with probability 0.01.

This means that
`p = 0.01`

The company sells the screws in packages of 10

This means that
`n = 10`

That is, the package would have to be replaced if greater or equal to 2 screws are defective. What proportion of packages sold must the company replace?

This is:

`P(X \geq 2) = 1 - P(X < 2)`

In which:

`P(X < 2) = P(X = 0) + P(X = 1)`

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

`P(X = 0) = C_(10,0).(0.01)^(0).(0.99)^(10) = 0.9044`

`P(X = 1) = C_(10,1).(0.01)^(1).(0.99)^(9) = 0.0914`

`P(X < 2) = P(X = 0) + P(X = 1) = 0.9044 + 0.0414 = 0.9958`

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9958 = 0.0042`

The company must replace 0.0042 = 0.42% of packages.