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,.....](https://tex.z-dn.net/?f=P%28X%3Dr%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7Dp%5E%7Br%7D%20%281-p%29%5E%7Bn-r%7D%20%3B%20x%20%3D%200%2C1%2C2%2C3%2C.....)
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%
<em>LET X = Number of U.S. mortgages those were delinquent in 2011</em>
So, it means X ~ ![Binom(n=8, p=0.08)](https://tex.z-dn.net/?f=Binom%28n%3D8%2C%20p%3D0.08%29)
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}](https://tex.z-dn.net/?f=%5Cbinom%7B8%7D%7B1%7D%5Ctimes%200.08%5E%7B1%7D%20%5Ctimes%20%281-0.08%29%5E%7B8-1%7D)
= ![8 \times 0.08 \times 0.92^{7}](https://tex.z-dn.net/?f=8%20%5Ctimes%200.08%20%5Ctimes%200.92%5E%7B7%7D)
= 0.357
<u><em>Hence, the probability that exactly one of these mortgages is delinquent is 0.357.</em></u>