Question

According to the Mortgage Bankers Association, 8% of U.S. mortgages were delinquent last year. A delinquent mortgage is one that has missed at least one payment but has not yet gone to foreclosure. A random sample of eight mortgages was selected. What is the probability that fewer than three of these mortgages are delinquent

Answer 1

0.9789 = 97.89% probability that fewer than three of these mortgages are delinquent.

Explanation:

For each mortgage, there are only two possible outcomes. Either it is delinquent, or it is not. The probability of a mortgage being delinquent is independent of any other mortgage. 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.

8% of U.S. mortgages were delinquent last year.

This means that
`p = 0.08`

A random sample of eight mortgages was selected.

This means that
`n = 8`

What is the probability that fewer than three of these mortgages are delinquent?

This is:

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

In which

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

`P(X = 0) = C_(8,0).(0.08)^(0).(0.92)^(8) = 0.5132`

`P(X = 1) = C_(8,1).(0.08)^(1).(0.92)^(7) = 0.3570`

`P(X = 2) = C_(8,2).(0.08)^(2).(0.92)^(6) = 0.1087`

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5132 + 0.3570 + 0.1087 = 0.9789`

0.9789 = 97.89% probability that fewer than three of these mortgages are delinquent.