/*! 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 59 At a certain gas station, \(40 \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At a certain gas station, \(40 \%\) of the customers 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

Identify Events and Probabilities

We are given three primary events related to gas type: regular \(A_1\), plus \(A_2\), and premium \(A_3\). The probabilities are \(P(A_1) = 0.40\), \(P(A_2) = 0.35\), and \(P(A_3) = 0.25\). Additionally, we have the conditional probabilities of filling the tank given the gas type: \(P(B|A_1) = 0.30\), \(P(B|A_2) = 0.60\), and \(P(B|A_3) = 0.50\).
02

Probability of Plus Gas and Filling the Tank

To find \(P(A_2 \cap B)\) (the probability of choosing plus gas and filling the tank), we use the multiplication rule: \((P(A_2 \cap B) = P(A_2) \times P(B|A_2))\). Substitute the values: \(P(A_2 \cap B) = 0.35 \times 0.60 = 0.21\).
03

Total Probability of Filling the Tank

To find \(P(B)\), the probability that a customer fills the tank, use the law of total probability: \((P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3))\). Substitute the values: \(P(B) = 0.30 \times 0.40 + 0.60 \times 0.35 + 0.50 \times 0.25 = 0.12 + 0.21 + 0.125 = 0.455\).
04

Conditional Probability for Each Gas Type Given the Tank is Filled

To find \(P(A_i|B)\) for each gas type, use Bayes' theorem: \((P(A_i|B) = \frac{P(B|A_i)P(A_i)}{P(B)})\). For regular gas: \(P(A_1|B) = \frac{0.30 \times 0.40}{0.455} = \frac{0.12}{0.455} \approx 0.2637\). For plus gas: \(P(A_2|B) = \frac{0.60 \times 0.35}{0.455} = \frac{0.21}{0.455} \approx 0.4615\). For premium gas: \(P(A_3|B) = \frac{0.50 \times 0.25}{0.455} = \frac{0.125}{0.455} \approx 0.2747\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 rule used to find the overall probability of an event that can occur in several ways. It helps in breaking down complex problems into simpler parts.
In our exercise, we wanted to find the probability that a customer fills the tank, regardless of the type of gas chosen. This requires considering all possible types of gas (regular, plus, and premium) and their individual probabilities of resulting in a filled tank.

We used the Law of Total Probability to calculate this total filling probability:
  • For regular gas: Multiply the probability of choosing regular gas by the probability of filling the tank using regular gas.
  • For plus gas: Multiply the probability of choosing plus gas by the probability of filling the tank using plus gas.
  • For premium gas: Multiply the probability of choosing premium gas by the probability of filling the tank using premium gas.
Finally, sum all these probabilities to find the overall probability that a customer fills the tank: \[ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3) \] This calculation helps in understanding how individual probabilities contribute to the total probability of an event.
Bayes' Theorem
Bayes' Theorem is a fascinating concept in probability theory that allows us to update our beliefs based on new information. It is particularly useful for calculating conditional probabilities in a reversed context.
In the problem, we wanted the probability of a customer requesting a specific type of gas, given that they filled the tank. This is where Bayes' Theorem comes into play.

To use Bayes' Theorem, we first calculate the probability of the customer filling the tank for each type of gas, as previously done with the Law of Total Probability. Then we apply Bayes' Theorem:
  • For regular gas, we find: \[ P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)} \]
  • For plus gas, we use: \[ P(A_2|B) = \frac{P(B|A_2)P(A_2)}{P(B)} \]
  • For premium gas, we calculate: \[ P(A_3|B) = \frac{P(B|A_3)P(A_3)}{P(B)} \]
Bayes' Theorem allows us to adjust our initial probability estimates conditional on new evidence, which in this case is filling the tank.
Multiplication Rule
The multiplication rule is a key part of probability theory, allowing us to find the probability of two events occurring together. This is also known as joint probability.
In the exercise, we needed to determine the probability of a customer choosing plus gas and filling the tank. This scenario involves two dependent events, requiring the use of the multiplication rule.

To find this joint probability, we carry out the following steps:
  • Identify the probability of selecting plus gas.
  • Identify the probability of filling the tank, given that the customer selected plus gas.
By using the multiplication rule, we combine these probabilities:\[ P(A_2 \cap B) = P(A_2) \times P(B|A_2) \]This calculation gives us the probability that both events (choosing plus gas and filling the tank) happen together.
Probability Theory
Probability theory is the backbone of understanding random events and how likely they are to occur. It provides the tools and rules, like Bayes' theorem and the Law of Total Probability, to calculate and understand probabilities.
In this exercise, probability theory allows us to predict customer behavior in choosing gas types and deciding whether to fill their tanks.

Key concepts include:
  • Understanding independent and dependent events.
  • Using conditional probability to update predictions based on new information.
  • Applying foundational rules, like the multiplication rule, to calculate complex joint probabilities.
We use probability theory to make sense of real-world scenarios, like decisions customers make at a gas station. By utilizing structured methods, we can make informed predictions and analysis in everyday situations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Four engineers, \(A, B, C\), and \(D\), have been scheduled for job interviews at \(10 \mathrm{~A} . \mathrm{M}\). on Friday, January 13 , at Random Sampling, Inc. The personnel manager has scheduled the four for interview rooms \(1,2,3\), and 4 , respectively. However, the manager's secretary does not know this, so assigns them to the four rooms in a completely random fashion (what else!). What is the probability that a. All four end up in the correct rooms? b. None of the four ends up in the correct room?

Show that if \(A_{1}, A_{2}\), and \(A_{3}\) are independent events, then \(P\left(A_{1} \mid A_{2} \cap A_{3}\right)=P\left(A_{1}\right)\) .

A subject is allowed a sequence of glimpses to detect a target. Let \(G_{i}=\\{\) the target is detected on the \(i\) th glimpse \(\\}\), with \(p_{i}=P\left(G_{i}\right)\). Suppose the \(G_{i}^{\prime}\) s are independent events, and write an expression for the probability that the target has been detected by the end of the \(n\)th glimpse. [Note: This model is discussed in "Predicting Aircraft Detectability" Human Factors, 1979: 277-291.]

As of April 2006, roughly 50 million .com web domain names were registered (e.g., yahoo.com). a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits as well as letters are permitted as characters? [Note: A character length of three or more is now mandated.] b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? [Note: All are currently taken.] c. Answer the questions posed in (b) for four-character sequences. d. As of April \(2006,97,786\) of the four-character sequences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?

A certain company sends \(40 \%\) of its overnight mail parcels via express mail service \(E_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (denote the event "late delivery" by \(L\) ). If a record of an overnight mailing is randomly selected from the company's file, what is the probability that the parcel went via \(E_{1}\) and was late?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.