Question

A real estate agent is showing homes to a prospective buyer. There are ten homes in the desired price range listed in the area. The buyer has time to visit only four of them. If four of the homes are new and six have previously been occupied and if the four homes to visit are randomly chosen, what is the probability that all four are new

Answer 1

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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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

$P(X = 4) = h(4,10,4,4) = \frac{C_{4,4}*C_{6,0}}{C_{10,4}} = 0.0048$

0.48% probability that all four are new