Question

Recent homebuyers from a local developer allege that 30% of the houses this developer constructs have some major defect that will require substantial repairs. To test this allegation, we randomly sample 20 homes constructed by the developer and find that two of the homes did indeed have some major defect. If the allegation is correct, what is the probability of observing at most two defective homes out of a random sample of 20

Answer 1

3.54% probability of observing at most two defective homes out of a random sample of 20

Explanation:

For each house that this developer constructs, there are only two possible outcomes. Either there are some major defect that will require substantial repairs, or there is not. The probability of a house having some major defect that will require substantial repairs is independent of other houses. 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.

30% of the houses this developer constructs have some major defect that will require substantial repairs.

This means that
`p = 0.3`

If the allegation is correct, what is the probability of observing at most two defective homes out of a random sample of 20

This is
`P(X \leq 2)` when n = 20. So

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)`

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

`P(X = 0) = C_(20,0).(0.3)^(0).(0.7)^(20) = 0.0008`

`P(X = 1) = C_(20,1).(0.3)^(1).(0.7)^(19) = 0.0068`

`P(X = 2) = C_(20,2).(0.3)^(2).(0.7)^(18) = 0.0278`

`P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0008 + 0.0068 + 0.0278 = 0.0354`

3.54% probability of observing at most two defective homes out of a random sample of 20