/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 A large operator of timeshare co... [FREE SOLUTION] | 91Ó°ÊÓ

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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 one-night 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?

Short Answer

Expert verified
Day visit: 0.087; One-night visit: 0.652; Two-night visit: 0.261.

Step by step solution

01

Define Events and Probabilities

We will define the relevant events and probabilities from the problem. Let \( D \) be the event that a visitor makes a day visit, \( N_1 \) be the event of a one-night visit, and \( N_2 \) for a two-night visit. Thus: \( P(D) = 0.20 \), \( P(N_1) = 0.50 \), \( P(N_2) = 0.30 \). We also know the conditional probabilities: \( P(Purchase|D) = 0.10 \), \( P(Purchase|N_1) = 0.30 \), and \( P(Purchase|N_2) = 0.20 \).
02

Calculate Total Probability of Purchase

Using the law of total probability, we need to find \( P(Purchase) \) which is the probability of making a purchase regardless of visit type. It is calculated as follows: \[P(Purchase) = P(Purchase|D) \cdot P(D) + P(Purchase|N_1) \cdot P(N_1) + P(Purchase|N_2) \cdot P(N_2) \] \[ = (0.10 \times 0.20) + (0.30 \times 0.50) + (0.20 \times 0.30) \] \[ = 0.02 + 0.15 + 0.06 = 0.23\].
03

Calculate Conditional Probabilities using Bayes' Theorem

We need to calculate \( P(D|Purchase) \), \( P(N_1|Purchase) \), and \( P(N_2|Purchase) \) using Bayes' theorem: \[P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\]For \( P(D|Purchase) \): \[P(D|Purchase) = \frac{P(Purchase|D) \cdot P(D)}{P(Purchase)} = \frac{0.10 \times 0.20}{0.23} = \frac{0.02}{0.23} \approx 0.087\] For \( P(N_1|Purchase) \):\[P(N_1|Purchase) = \frac{P(Purchase|N_1) \cdot P(N_1)}{P(Purchase)} = \frac{0.30 \times 0.50}{0.23} = \frac{0.15}{0.23} \approx 0.652\] For \( P(N_2|Purchase) \):\[P(N_2|Purchase) = \frac{P(Purchase|N_2) \cdot P(N_2)}{P(Purchase)} = \frac{0.20 \times 0.30}{0.23} = \frac{0.06}{0.23} \approx 0.261\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conditional Probability
Conditional probability is a fundamental concept in probability theory that measures the likelihood of an event occurring, given that another event has already happened. In our exercise, this is represented by scenarios such as knowing a person made a purchase and then asking how likely it is that they visited for a specific number of nights.

To put it simply, conditional probability helps us update our knowledge about the probability of an event after taking into account new information. For instance, we know the general probability of someone purchasing after a one-night visit, which is different from just knowing the general probability of a one-night visit.

In mathematical terms, the conditional probability of event A given event B is denoted as \(P(A|B)\) and is calculated using Bayes' theorem as follows:
  • \(P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\)
With this formula, we can effectively "invert" conditions, as we did in the exercise when looking for the probability of different visit durations given that a purchase was made.
Law of Total Probability
The law of total probability is a rule that provides a way to calculate the total probability of an event based on its registration across multiple disjoint events. It enables us to partition the probability space into simpler discrete probabilities associated with these events.

For our scenario, think of it as examining all different visitor types (day, one-night, two-night visits) and how each contributes to the total probability of making a purchase regardless of the type. This not only organizes our calculations but ensures that we consider all possibilities linked to an outcome.
Formally, the law of total probability is expressed as:
  • \(P(B) = P(B|A_1) \cdot P(A_1) + P(B|A_2) \cdot P(A_2) + \ldots + P(B|A_n) \cdot P(A_n)\)
Here, each \( A_i \) corresponds to different independent events, just like different visitor options in our case. By calculating it, we account for every scenario – from visiting for a day to staying overnight – and how likely it leads to purchasing overall.
Probability Theory
Probability theory forms the backbone of statistics and data analysis and is essential in understanding the likelihood of various outcomes in uncertain contexts. It involves several concepts like events, random variables, and probability distributions, which help in measuring uncertainty and risk.

In the context of our problem, probability theory helps us deal with the uncertainty of whether a person makes a purchase, given different visiting durations. We use probabilities not only to describe individual outcomes but also the relationship between those outcomes through conditional probabilities and the law of total probability.
Probability theory equips us with the tools to delve deeper into understanding events and their potential outcomes. It allows us to see how the initial probability of individual events (like choosing a visit duration) combined with conditional probabilities (choosing ones who purchased) culminates in specific calculated outcomes, like knowing the conditional probabilities for each visitor type.
  • In essence, probability theory is about quantifying randomness and predicting patterns, an invaluable skill in decision-making and data-driven insights.

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Most popular questions from this chapter

A computer consulting firm presently has bids out on three projects. Let \(A_{i}=\\{\) awarded project \(i\\}\), for \(i=1,2,3\), and suppose that \(P\left(A_{1}\right)=.22, P\left(A_{2}\right)=.25, P\left(A_{3}\right)=.28\), \(P\left(A_{1} \cap A_{2}\right)=.11, P\left(A_{1} \cap A_{3}\right)=.05, P\left(A_{2} \cap A_{3}\right)=.07\), \(P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01\). Express in words each of the following events, and compute the probability of each event: a. \(A_{1} \cup A_{2}\) b. \(A_{1}^{\prime} \cap A_{2}^{\prime}\) c. \(A_{1} \cup A_{2} \cup A_{3}\) d. \(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}^{\prime}\) e. \(A_{1}^{\prime} \cap A_{2}^{\prime} \cap A_{3}\) f. \(\left(A_{1}^{\prime} \cap A_{2}^{\prime}\right) \cup A_{3}\)

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries. a. If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this? b. If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this? c. If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? d. If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen? e. If 6 bottles are randomly selected, what is the probability that all of them are the same variety?

Let \(A\) denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let \(B\) be the event that the next request is for help with SAS. Suppose that \(P(A)=.30\) and \(P(B)=.50\). a. Why is it not the case that \(P(A)+P(B)=1\) ? b. Calculate \(P\left(A^{\prime}\right)\). c. Calculate \(P(A \cup B)\). d. Calculate \(P\left(A^{\prime} \cap B^{\prime}\right)\).

Again consider a Little League team that has 15 players on its roster. a. How many ways are there to select 9 players for the starting lineup? b. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters? c. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players?

Suppose that \(55 \%\) of all adults regularly consume coffee, \(45 \%\) regularly consume carbonated soda, and \(70 \%\) regularly consume at least one of these two products. a. What is the probability that a randomly selected adult regularly consumes both coffee and soda? b. What is the probability that a randomly selected adult doesn't regularly consume at least one of these two products?

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