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At a certain gas station, \(40 \%\) of the castomers use regular gas \(\left(A_{1}\right), 35 \%\) use plus gas \(\left(A_{2}\right)\), and \(25 \%\) use premium \(\left(A_{3}\right)\). Of those customers using regular gas, only \(30 \%\) fill their tanks (event \(B\) ). Of those customers using plus, \(60 \%\) fill their tanks, whereas of those using premium, \(50 \%\) fill their tanks. a. What is the probability that the next customer will request plus gas and fill the tank \(\left(A_{2} \cap B\right)\) ? b. What is the probability that the next customer fills the tank? c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?

Short Answer

Expert verified
a. 0.21 b. 0.455 c. Regular: 0.2637, Plus: 0.4615, Premium: 0.2747

Step by step solution

01

Calculate Probability of Event A2 and B

The event where a customer requests plus gas and fills the tank is denoted by \( A_2 \cap B \). The probability of this event can be calculated by multiplying the probability of requesting plus gas with the probability of filling the tank after requesting plus gas. Thus, we have:\[ P(A_2 \cap B) = P(A_2) \times P(B|A_2) \]Given, \( P(A_2) = 0.35 \) and \( P(B|A_2) = 0.60 \), so:\[ P(A_2 \cap B) = 0.35 \times 0.60 = 0.21 \]
02

Calculate Total Probability of Filling the Tank (Event B)

To find the probability that a customer fills the tank, we use the law of total probability. We need to add up the probabilities of filling the tank from each gas type:\[ P(B) = P(A_1) \times P(B|A_1) + P(A_2) \times P(B|A_2) + P(A_3) \times P(B|A_3) \]Plug in the given probabilities:\[ P(B) = 0.40 \times 0.30 + 0.35 \times 0.60 + 0.25 \times 0.50 \]Calculate each term:\[ P(B) = 0.12 + 0.21 + 0.125 = 0.455 \]
03

Calculate Conditional Probability of Requesting Each Gas Type Given They Fill the Tank

For part c, we need to determine the conditional probability of each gas type given that the tank is filled, using Bayes' theorem:\[ P(A_1|B) = \frac{P(A_1 \cap B)}{P(B)}, \quad P(A_2|B) = \frac{P(A_2 \cap B)}{P(B)}, \quad P(A_3|B) = \frac{P(A_3 \cap B)}{P(B)} \]Calculate for each gas type:\[ P(A_1 \cap B) = 0.40 \times 0.30 = 0.12 \]\[ P(A_2 \cap B) = 0.35 \times 0.60 = 0.21 \]\[ P(A_3 \cap B) = 0.25 \times 0.50 = 0.125 \]For regular gas:\[ P(A_1|B) = \frac{0.12}{0.455} \approx 0.2637 \]For plus gas:\[ P(A_2|B) = \frac{0.21}{0.455} \approx 0.4615 \]For premium gas:\[ P(A_3|B) = \frac{0.125}{0.455} \approx 0.2747 \]

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Key Concepts

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

Law of Total Probability
The Law of Total Probability is a fundamental principle in probability theory that helps us calculate the total probability of an event by considering all possible ways in which the event can occur. In this context, the law was used to find the probability that a customer fills their tank, irrespective of the type of gas they choose.
To apply the law, we first identify all the different scenarios in which a customer could fill their tank:
  • Using regular gas and filling the tank.
  • Using plus gas and filling the tank.
  • Using premium gas and filling the tank.
To find the total probability of a customer filling their tank (\( P(B) \)), we sum up the probabilities of these individual events:\[P(B) = P(A_1) \times P(B|A_1) + P(A_2) \times P(B|A_2) + P(A_3) \times P(B|A_3)\]Breaking it down using given probabilities, we get:\[0.40 \times 0.30 + 0.35 \times 0.60 + 0.25 \times 0.50 = 0.455\]Therefore, there's a \(45.5\% \) chance that a customer will fill their tank.
Bayes' Theorem
Bayes' Theorem provides us with a way to update the probability of an event based on new evidence. It's a powerful tool for understanding conditional probability. In this exercise, Bayes' Theorem was employed to determine the likelihood of each type of gas being requested, given that the tank is filled.
The theorem is expressed as:
  • \( P(A|B) = \frac{P(A \cap B)}{P(B)} \)
Here, \( P(A|B) \) is the conditional probability of event \( A \) given event \( B \), \( P(A \cap B) \) is the joint probability of both events occurring, and \( P(B) \) is the probability of the new evidence.For this problem, we calculated:
  • For regular gas (\( A_1 \)): \( P(A_1|B) = \frac{0.12}{0.455} \approx 0.2637 \), meaning if a customer fills the tank, there is about a 26.37% chance they requested regular gas.
  • For plus gas (\( A_2 \)): \( P(A_2|B) = \frac{0.21}{0.455} \approx 0.4615 \), indicating a 46.15% chance.
  • For premium gas (\( A_3 \)): \( P(A_3|B) = \frac{0.125}{0.455} \approx 0.2747 \), reflecting a 27.47% chance.
This illustrates how Bayes' Theorem helps refine predictions based on observed data.
Joint Probability
Joint Probability refers to the likelihood of two events happening at the same time. In probability calculations, it often involves events that can simultaneously occur. In this exercise, we're particularly interested in the scenario where a customer both chooses a type of gas and fills the tank.
Joint probability is typically calculated by multiplying the probability of the first event by the conditional probability of the second event given the first:\[P(A \cap B) = P(A) \times P(B|A)\]For example, considering plus gas and filling the tank (\( A_2 \cap B \)), we calculated the joint probability as follows:
  • Start with the probability that a customer opts for plus gas (\( P(A_2) = 0.35 \)).
  • Multiply that by the probability that they fill the tank, given they chose plus gas (\( P(B|A_2) = 0.60 \)).
Thus:\[P(A_2 \cap B) = 0.35 \times 0.60 = 0.21\]This means there is a 21% probability that any given customer will both request plus gas and fill their tank. Joint probability is crucial in assessing the likelihood of combined events.

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Most popular questions from this chapter

Consider the following information about travelers on vacation (based partly on a recent Travelocity poll): \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them, \(23 \%\) both check work e-mail and use a cell phone to stay connected, and \(51 \%\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 88 out of every 100 who bring a laptop also check work email, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop. a. What is the probability that a randomly selected traveler who checks work email also uses a cell phone to stay connected? b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? c. If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/she uses a cell phone to stay connected?

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.

Show that for any three events \(A, B\), and \(C\) with \(P(C)>0\), \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\).

Show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B\) ), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B)\), \(P(A)\), and \(P(A \cup B)\) ?

For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+\) \(P\left(A^{\prime} \mid B\right)=1\).

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