Question

A large operator of timeshare complexes requires anyone interested in making a purchase to first visit the site of interest. Historical data indicates that 20% of all potential purchasers select a day visit, 50% choose a one-night visit, and 30% opt for a two-night visit. In addition, 10% of day visitors ultimately make a purchase, 30% of onenight visitors buy a unit, and 20% of those visiting for two nights decide to buy. Suppose a visitor is randomly selected and is found to have made a purchase. How likely is it that this person made a day visit? A one-night visit? A two-night visit?

Answer 1

Answer:

0.087 = 8.7% probability that this person made a day visit.

0.652 = 65.2% probability that this person made a one-night visit.

0.261 = 26.1% probability that this person made a two-night visit.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

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

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 question:

Event A: Made a purchase.

Probability of making a purchase:

10% of 20%(day visit)

30% of 50%(one night)

20% of 30%(two night).

So

$p = 0.1*0.2 + 0.3*0.5 + 0.2*0.3 = 0.23$

How likely is it that this person made a day visit?

Here event B is a day visit.

10% of 20% is the percentage of purchases and day visit. So

$P(A \cap B) = 0.1*0.2 = 0.02$

So

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.02}{0.23} = 0.087$

0.087 = 8.7% probability that this person made a day visit.

A one-night visit?

Event B is a one night visit.

The percentage of both(one night visit and purchase) is 30% of 50%. So

$P(A \cap B) = 0.3*0.5 = 0.15$

So

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.15}{0.23} = 0.652$

0.652 = 65.2% probability that this person made a one-night visit.

A two-night visit?

Event B is a two night visit.

The percentage of both(two night visit and purchase) is 20% of 30%. So

$P(A \cap B) = 0.2*0.3 = 0.06$

Then

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.23} = 0.261$

0.261 = 26.1% probability that this person made a two-night visit.