Question

A real estate agent has 17 properties that she shows. She feels that there is a 50% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling less than 3 properties in one week. Round your answer to four decimal places.

Answer 1

Answer:

0.0011 = 0.11% probability of selling less than 3 properties in one week.

Step-by-step explanation:

For each property, there are only two possible outcomes. Either they are sold, or they are not. The chance of selling any one property is independent of selling another property. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

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

And p is the probability of X happening.

A real estate agent has 17 properties that she shows.

This means that $n = 17$

She feels that there is a 50% chance of selling any one property during a week.

This means that $p = 0.5$

Compute the probability of selling less than 3 properties in one week.

This is

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{17,0}.(0.5)^{0}.(0.5)^{17} \approx 0$

$P(X = 1) = C_{17,1}.(0.5)^{1}.(0.5)^{16} = 0.0001$

$P(X = 2) = C_{17,2}.(0.5)^{2}.(0.5)^{15} = 0.0010$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.0001 + 0.0010 = 0.0011$

0.0011 = 0.11% probability of selling less than 3 properties in one week.