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, 30% of day visitors ultimately make a purchase, 10% of one-night visitors buy a unit, and 50% 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

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joaobezerra joaobezerra · Answer 1 · 2021-03-08T01:15:02+01:00

Answer:

0.2308 = 23.08% probability that this person made a day visit

Step-by-step explanation:

Conditional Probability

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

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

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

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Visitor made a purchase

Event B: Made a day visit

Probability that a visitor made a purchase:

30% of 20%(day visitors)

10% of 50%(one-night)

50% of 30%(two-night)

So

[tex]P(A) = 0.3*0.2 + 0.1*0.5 + 0.5*0.3 = 0.26[/tex]

Probability of a purchase with a day visit:

30% of 20%. So

[tex]P(A \cap B) = 0.3*0.2 = 0.06[/tex]

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

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.06}{0.26} = 0.2308[/tex]

0.2308 = 23.08% probability that this person made a day visit