Mathematics, 10.10.2019 00:00 bcarri4073

Alarge 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, 40% of day visitors ultimately make a purchase, 30% 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? (round your answer to three decimal places.) how likely is it that this person made a one-night visit? (round your answer to three decimal places.) how likely is it that this person made a two-night visit? (round your answer to three decimal places.)

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Answer from: goldenarrow

There is a 21.053% probability that this person made a day visit.

There is a 39.474% probability that this person made a one night visit.

There is a 39.474% probability that this person made a two night visit.

Step-by-step explanation:

We have these following percentages

20% select a day visit

50% select a one-night visit

30% select a two-night visit

40% of the day visitors make a purchase

30% of one night visitors make a purchase

50% of two night visitors make a purchase

The first step to solve this problem is finding the probability that a randomly selected visitor makes a purchase. So:

P = 0.2(0.4) + 0.5(0.3) + 0.3(0.5) = 0.38

There is a 38% probability that a randomly selected visitor makes a purchase.

Now, as for the questions, we can formulate them as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

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?

What is the probability that this person made a day visit, given that she made a purchase?

P(B) is the probability that the person made a day visit. So P(B) = 0.20

P(A/B) is the probability that the person who made a day visit made a purchase. So P(A/B) = 0.4

P(A) is the probability that the person made a purchase. So P(A) = 0.38

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.4*0.2}{0.38} = 0.21053$

There is a 21.053% probability that this person made a day visit.

How likely is it that this person made a one-night visit?

What is the probability that this person made a one night visit, given that she made a purchase?

P(B) is the probability that the person made a one night visit. So P(B) = 0.50

P(A/B) is the probability that the person who made a one night visit made a purchase. So P(A/B) = 0.3

P(A) is the probability that the person made a purchase. So P(A) = 0.38

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.5*0.3}{0.38} = 0.39474$

There is a 39.474% probability that this person made a one night visit.

How likely is it that this person made a two-night visit?

What is the probability that this person made a two night visit, given that she made a purchase?

P(B) is the probability that the person made a two night visit. So P(B) = 0.30

P(A/B) is the probability that the person who made a two night visit made a purchase. So P(A/B) = 0.5

P(A) is the probability that the person made a purchase. So P(A) = 0.38

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.3*0.5}{0.38} = 0.39474$

There is a 39.474% probability that this person made a two night visit.

Answer from: Quest

it depends on the weather. because i have flown to seattle when it was pouring down rain and there was one time i flew to ohio when it was snowing hard. whats your situation.

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

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