Answer:
0.48% probability that all four are new
Step-by-step explanation:
The homes are chosen "without replacement", which means that after a home is visited, it is not elegible to be visited again. So we use the hypergeometric distribution to solve this question.
Hypergeometric distribution:
The probability of x sucesses is given by the following formula:
![P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20h%28x%2CN%2Cn%2Ck%29%20%3D%20%5Cfrac%7BC_%7Bk%2Cx%7D%2AC_%7BN-k%2Cn-x%7D%7D%7BC_%7BN%2Cn%7D%7D)
In which:
x is the number of sucesses.
N is the size of the population.
n is the size of the sample.
k is the total number of desired outcomes.
Combinations formula:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
In this question:
Total of 10 homes, so N = 10.
We want 4 new, so x = 4.
In total, there are 4 new, so k = 4.
Sample of four homes, so n = 4.
Then
![P(X = 4) = h(4,10,4,4) = \frac{C_{4,4}*C_{6,0}}{C_{10,4}} = 0.0048](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20h%284%2C10%2C4%2C4%29%20%3D%20%5Cfrac%7BC_%7B4%2C4%7D%2AC_%7B6%2C0%7D%7D%7BC_%7B10%2C4%7D%7D%20%3D%200.0048)
0.48% probability that all four are new