Question

According to the Mortgage Bankers Association, 8% of U.S. mortgages were delinquent in 2011. 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 exactly one of these mortgages is delinquent?

Answer 1

Answer:

The probability that exactly one of these mortgages is delinquent is 0.357.

Step-by-step explanation:

We are given that according to the Mortgage Bankers Association, 8% of U.S. mortgages were delinquent in 2011. 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.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 mortgages

            r = number of success = exactly one

           p = probability of success which in our question is % of U.S.

                  mortgages those were delinquent in 2011, i.e; 8%

LET X = Number of U.S. mortgages those were delinquent in 2011

So, it means X ~ $Binom(n=8, p=0.08)$

Now, Probability that exactly one of these mortgages is delinquent is given by = P(X = 1)

                 P(X = 1)  = $\binom{8}{1}\times 0.08^{1} \times (1-0.08)^{8-1}$

                               = $8 \times 0.08 \times 0.92^{7}$

                               = 0.357

Hence, the probability that exactly one of these mortgages is delinquent is 0.357.