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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

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.14\) and \(P(F)=0.76\) i. \(\quad\) What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F)\) ? b. Suppose that for events \(A\) and \(B, P(A)=0.24, P(B)=0.24\) and \(P(A \cup B)=0.48\). Are \(A\) and \(B\) mutually exclusive? How can you tell?

Suppose that an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is over 6 feet in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is greater, \(P(A \mid B)\) or \(P(B \mid A) ?\) Why?

Consider a chance experiment that consists of selecting a customer at random from all people who purchased a car at a large car dealership during 2016 . a. In the context of this chance experiment, give an example of two events that would be mutually exclusive. b. In the context of this chance experiment, give an example of two events that would not be mutually exclusive.

Lyme disease is transmitted by infected ticks. Several tests are available for people with symptoms of Lyme disease. One of these tests is the EIA/IFA test. The paper "Lyme Disease Testing by Large Commercial Laboratories in the United States" (Clinical Infectious Disease [2014]: \(676-681\) ) found that \(11.4 \%\) of those tested actually had Lyme disease. Consider the following events: \(+\) represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities are based on percentages given in the paper: $$ \begin{array}{r} P(L)=0.114 \\ P\left(L^{C}\right)=0.886 \end{array} $$ $$ \begin{array}{c} P(+\mid L)=0.933 \\ P(-\mid L)=0.067 \\ P\left(+\mid L^{C}\right)=0.039 \\ P\left(-\mid L^{C}\right)=0.961 \end{array} $$ a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a hypothetical 1000 table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use information from the table constructed in Part (b) to calculate this probability.

The article "Scrambled Statistics: What Are the Chances of Finding Multi-Yolk Eggs?" (Significance [August 2016]: 11) gives the probability of a double-yolk egg as 0.001 . a. Give a relative frequency interpretation of this probability. b. If 5000 eggs were randomly selected, about how many double-yolk eggs would you expect to find?
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