Question

Alison has been hired to sell two different homes on the same street that two houses apart. She predicts that Home A has a 75% chance in selling on the first week of being listed, whereas Home B is in lesser condition and has a 30% probability. There is also a 20% chance both homes will not sell on the first week of it being listed. What is the probability that house A does not sell given that house B does not sell due to it's poor condition

Answer 1

Using conditional probability, it is found that there is a 0.2857 = 28.57% probability that house A does not sell given that house B does not sell due to it's poor condition.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this problem:

• Event A: House B does not sell.
• Event B: House A does not sell.

Home B has a 30% probability of selling, so a 100 - 30 = 70% probability of not selling, hence $P(A) = 0.7$.

20% probability that none sells, hence $P(A \cap B) = 0.2$.

Then, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.2}{0.7} = 0.2857$

0.2857 = 28.57% probability that house A does not sell given that house B does not sell due to it's poor condition.

A similar problem is given at brainly.com/question/14398287