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Chapter 5: Problem 37

A rental car company offers two options when a car is rented. A renter can choose to pre-purchase gas or not and can also choose to rent a GPS device or not. Suppose that the events \(A=\) event that gas is pre-purchased \(B=\) event that a GPS is rented are independent with \(P(A)=0.20\) and \(P(B)=0.15\). a. Construct a hypothetical 1000 table with columns corresponding to whether or not gas is pre-purchased and rows corresponding to whether or not a GPS is rented. b. Use the table to find \(P(A \cup B)\). Give a long-run relative frequency interpretation of this probability.

Short Answer

Expert verified
In this problem, we have independent events \(A\) - pre-purchasing gas and \(B\) - renting a GPS device, with probabilities \(P(A)=0.20\) and \(P(B)=0.15\). We constructed a hypothetical table of 1000 cases presenting combinations of these events and calculated the probability of the union of events A and B as follows: \(P(A \cup B) = \frac{30 + 120 + 170}{1000} = \frac{320}{1000} = 0.32\) This means that in the long run, for 1000 car rentals, about 32% (or 320 out of 1000 times) the renter would either pre-purchase gas or rent a GPS device or both.

Step by step solution

01

Understanding the given information

We know that events \(A\) and \(B\) are independent, meaning that the occurrence of one event doesn't affect the occurrence of the other event. We are given the probabilities: \(P(A) = 0.20\) and \(P(B) = 0.15\). To calculate \(P(A \cup B)\) for independent events, we can use the following formula: \(P(A \cup B) = P(A) + P(B) - P(A)P(B)\)
02

Deriving probabilities from the given information

Since events A and B are independent, we can compute probabilities of their complements as follows: 1. Probability of not pre-purchasing gas: \(P(A') = 1 - P(A) = 1 - 0.20 = 0.80\) 2. Probability of not renting a GPS: \(P(B') = 1 - P(B) = 1 - 0.15 = 0.85\)
03

Constructing the table using the hypothetical 1000 cases

To construct the table, we will use the probabilities derived in Step 2 and multiply each with the hypothetical 1000 cases. The table will look like this: | | Gas Pre-Purchased (A) | Gas Not Pre-Purchased (A') | Total | |----------------|-----------------------|----------------------------|-------| | GPS Rented (B) | 30 | 120 | 150 | | No GPS (B') | 170 | 680 | 850 | | Total | 200 | 800 | 1000 | The values in the table are calculated using the probabilities and the hypothetical 1000 cases, e.g., -GPS rented (B) and gas pre-purchased (A): \(1000 \times P(A) \times P(B) = 1000 \times 0.20 \times 0.15 = 30\) -GPS rented (B) and gas not pre-purchased (A'): \(1000 \times P(A') \times P(B) = 1000 \times 0.80 \times 0.15 = 120\) -And other values are calculated in a similar manner.
04

Calculate the probability of the union of events A and B

Using the values from the table, we can calculate the probability of the union of events A and B as follows: \(P(A \cup B) = \frac{30 + 120 + 170}{1000} = \frac{320}{1000} = 0.32\) The long-run relative frequency interpretation of this probability is that, in the long run, for 1000 car rentals, about 32% (or 320 out of 1000 times) the renter would either pre-purchase gas or rent a GPS device or both.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that helps us understand how likely events are to occur. It provides us with tools to quantify uncertainty and make informed predictions based on the likelihood of different outcomes.

There are some key concepts involved in probability theory:
  • **Probability Mass Function (PMF):** Deals with discrete random variables, where each possible outcome has an associated probability.
  • **Probability Density Functions (PDF):** Used for continuous random variables, showing distribution over a range.
  • **Events:** Outcomes or combinations of outcomes within a sample space.
  • **Sample Space:** The set of all possible outcomes.
In our exercise, we deal with two independent events: pre-purchasing gas and renting a GPS, where probabilities like \(P(A) = 0.20\) and \(P(B) = 0.15\) illustrate how often these events occur.
Long-Run Relative Frequency
Long-run relative frequency is a concept in probability that explains how often an event occurs over many trials or occurrences. It is the proportion of times that a particular event happens when an experiment is repeated a large number of times.

For example, in our rental car exercise, the probability of a union, \(P(A \cup B) = 0.32\), means that in the long run, about 32% of the time, either gas is pre-purchased, a GPS is rented, or both occur. In a setting of 1000 rentals, this probability suggests that 320 rentals would involve at least one of these actions.
  • This concept helps to demonstrate how probability works in real life.
  • It translates theoretical probabilities into expected occurrences over time.
Probability of Union
The probability of the union of two events, denoted \(P(A \cup B)\), refers to the chance of at least one of the events happening. The formula we often use for this calculation is:
  • \(P(A \cup B) = P(A) + P(B) - P(A)P(B)\)
This formula is especially useful when the two events A and B are independent. It helps avoid double counting the scenario where both events happen simultaneously.

In our scenario, they calculated \(P(A \cup B) = 0.32\), indicating that there's a 32% chance that either one or both events occur in a rental car situation. This probability tells us how likely it is to have either independent event, expanding our understanding of possible outcomes.
Complementary Events
Complementary events are pairs of outcomes where only one can occur at a time, and together they account for all possibilities. If the event is \(A\), then its complement is \(A'\), representing all outcomes not in \(A\).

Mathematically, the sum of an event's probability and its complement is always 1:
  • \(P(A) + P(A') = 1\)
In our exercise, we have:
  • \(P(A') = 1 - P(A) = 0.80\)
  • \(P(B') = 1 - P(B) = 0.85\)
Understanding complementary events helps determine the likelihood of both occurrence and non-occurrence, providing a complete picture of the situation's probabilities.

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Most popular questions from this chapter

In a small city, approximately \(15 \%\) of those eligible are called for jury duty in any one calendar year. People are selected for jury duty at random from those eligible, and the same individual cannot be called more than once in the same year. What is the probability that an eligible person in this city is selected in both of the next 2 years? All of the next 3 years?

Suppose you want to estimate the probability that a randomly selected customer at a particular grocery store will pay by credit card. Over the past 3 months, 80,500 purchases were made, and 37,100 of them were paid for by credit card. What is the estimated probability that a randomly selected customer will pay by credit card?

5.62 An appliance manufacturer offers extended warranties on its washers and dryers. Based on past sales, the manufacturer reports that of customers buying both a washer and a dryer, \(52 \%\) purchase the extended warranty for the washer, \(47 \%\) purchase the extended warranty for the dryer, and \(59 \%\) purchase at least one of the two extended warranties. In Exercise \(5.34,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that a randomly selected customer who buys a washer and a dryer purchases an extended warranty for both the washer and the dryer. b. The probability that a randomly selected customer does not purchase an extended warranty for either the washer or dryer.

What does it mean to say that the probability that a coin toss will land head side up is \(0.5 ?\)

A professor assigns five problems to be completed as homework. At the next class meeting, two of the five problems will be selected at random and collected for grading. You have only completed the first three problems. a. What is the sample space for the chance experiment of selecting two problems at random? (Hint: You can think of the problems as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E}\). One possible selection of two problems is \(\mathrm{A}\) and \(\mathrm{B}\). If these two problems are selected and you did problems \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\), you will be able to turn in both problems. There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that you will be able to turn in both of the problems selected? d. Does the probability that you will be able to turn in both problems change if you had completed the last three problems instead of the first three problems? Explain. e. What happens to the probability that you will be able to turn in both problems selected if you had completed four of the problems rather than just three?
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