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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
The probability of either pre-purchasing gas or renting a GPS device or both is \(0.32\) or 32%, which in a scenario of 1000 rentals corresponds to 320 rentals.

Step by step solution

01

Hypothetical 1000 table

To start with the construction of the hypothetical 1000 rentals table, calculate the number of times each event happens in these rentals. Given that \(P(A)=0.20\) and \(P(B)=0.15\), we multiply each probability by 1000 to obtain:- Pre-purchased gas (event A): \(0.20 * 1000 = 200\)- Rented GPS (event B) : \(0.15 * 1000 = 150\)So, in a hypothetical 1000 rentals scenario, gas would be pre-purchased 200 times and a GPS would be rented 150 times.
02

Calculate the intersections

Next, to complete the hypothetical 1000 table, we also need to know how many times both events A and B occur simultaneously. Since the events are independent, the joint probability is simply the product of their individual probabilities. Therefore, \(P(A \cap B) = P(A)P(B) = 0.20 * 0.15 = 0.03\). To get this in terms of a hypothetical 1000, multiply by 1000: \(0.03 * 1000 = 30\). So, in our hypothetical 1000 rentals scenario, both gas is pre-purchased and a GPS is rented 30 times in total.
03

Calculate \(P(A \cup B)\)

The probability of the union of events A and B is given by the formula \(P(A \cup B) = P(A) + P(B) - P(A)P(B)\). Substituting given probabilities into the formula, we get: \(P(A \cup B) = 0.20 + 0.15 - 0.20*0.15 = 0.32 \). This is the long-run relative frequency for either pre-purchasing gas, renting 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.

Understanding the Hypothetical 1000 Table
When faced with understanding probabilities for different events, visual aids can often shed more light than abstract figures. This is where the hypothetical 1000 table comes in handy. It's a simple yet effective way to grasp how probabilities play out over a larger number of trials.

Imagine setting up a table with 1000 rows, where each row represents a rental transaction. By multiplying the event probabilities with 1000, we can visualize these probabilities as actual counts out of 1000. If the probability of pre-purchasing gas (event A) is 0.20, we'd expect to see pre-purchased gas in about 200 of these hypothetical rentals. Similarly, for renting a GPS (event B) with a probability of 0.15, we'd see a GPS rented in approximately 150 cases.

This kind of table makes it easier to comprehend the scale of probability, and more importantly, the interplay between different events, especially when you're digging into concepts like joint and union probabilities. It's a concrete way of seeing abstract relationships, which is crucial for those who need to visualize concepts to understand them better.
Joint Probability Explained
Joint probability is a core concept in understanding how two events coincide with each other. In our exercise, we examine the joint probability of two independent events: pre-purchasing gas and renting a GPS.

Events are independent if the occurrence of one does not affect the probability of the occurrence of the other. Calculating the joint probability of two independent events is fairly straight-forward; we simply multiply the probabilities of each event occurring separately. Mathematically, it's expressed as:\[ P(A \cap B) = P(A) \times P(B) \].

For our hypothetical rentals, with probabilities of 0.20 and 0.15 respectively, the joint probability becomes 0.03. This means, if you were to look at these hypothetical 1000 rentals, around 30 would have both gas pre-purchased and a GPS rented. This joint probability is a crucial piece in determining the overall landscape of decisions made by the rental car company's customers, helping them to strategize their offers and marketing.
Union of Events Probability and Its Importance
Lastly, we arrive at the concept of the union of events probability. This is the likelihood that at least one of two events will occur. In this context, we're referring to either pre-purchasing gas or renting a GPS—or both happening together.

The formula to calculate the union of two events A and B is:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \].

Using this formula, we subtract the joint probability from the sum of the probabilities of events A and B to avoid double-counting the cases where both events occur. For our exercise, the union probability turns out to be 0.32. Interpretatively, it denotes that out of 1000 rentals, we would expect 320 to involve either pre-purchasing gas, renting a GPS device, or both. This concept is essential for businesses and statisticians to assess the range of customer behaviors and to make informed decisions or predictions based on customer tendencies.

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Most popular questions from this chapter

The paper "Action Bias among Elite Soccer Goalkeepers: The Case of Penalty Kicks" ( Journal of Economic Psychology [2007]: \(606-621\) ) presents an interesting analysis of 286 penalty kicks in televised championship soccer games from around the world. In a penalty kick, the only players involved are the kicker and the goalkeeper from the opposing team. The kicker tries to kick a ball into the goal from a point located 11 meters away. The goalkeeper tries to block the ball from entering the goal. For each penalty kick analyzed, the researchers recorded the direction that the goalkeeper moved (jumped to the left, stayed in the center, or jumped to the right) and whether or not the penalty kick was successfully blocked. Consider the following events: \(L=\) the event that the goalkeeper jumps to the left \(C=\) the event that the goalkeeper stays in the center \(R=\) the event that the goalkeeper jumps to the right \(B=\) the event that the penalty kick is blocked Based on their analysis of the penalty kicks, the authors of the paper gave the following probability estimates: $$ \begin{array}{rrr} P(L)=0.493 & P(C)=0.063 & P(R)=0.444 \\ P(B \mid L)=0.142 & P(B \mid C)=0.333 & P(B \mid R)=0.126 \end{array} $$ 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 penalty kick was blocked and rows corresponding to whether the goalkeeper jumped left, stayed in the center, or jumped right. (Hint: See Example 5.14) c. Use the table to calculate the probability that a penalty kick is blocked. d. Based on the given probabilities and the probability calculated in Part (c), what would you recommend to a goalkeeper as the best strategy when trying to defend against a penalty kick? How does this compare to what goalkeepers actually do when defending against a penalty kick?

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