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?

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

In which

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

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

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

So

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

So

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

Then

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

ok

step-by-step explanation:

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide