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 three of them.

If four homes are new and six have previously been occupied and if the three homes to visit are randomly chosen, what is the probability that all three are new? (The same answer results regardless of whether order is considered.)

Answer 1

The probability that all three homes selected are new is 1/30.

Step-by-step explanation:

The probability that all three homes are new when randomly choosing three from a total of ten homes (four new, six previously occupied) can be calculated using combinations. The number of combinations of selecting 3 new homes from 4 is given by C(4,3), which is the number of ways to choose 3 from 4 without regard to order. Similarly, the total number of combinations of selecting any 3 homes from 10 is given by C(10,3).

To find the required probability, we use the formula:

Probability = (Number of favorable outcomes) / (Total number of outcomes) = C(4,3) / C(10,3)

Using the combination formula:

C(n,r) = n! / (r!(n-r)!)

We calculate C(4,3) and C(10,3):

• C(4,3) = 4! / (3!(4-3)!) = 4
• C(10,3) = 10! / (3!(10-3)!) = 120

So, the probability is:

P(all three homes are new) = 4 / 120 = 1 / 30