/*! 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 53 A certain shop repairs both audi... [FREE SOLUTION] | 91Ó°ÊÓ

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A certain shop repairs both audio and video components. Let \(A\) denote the event that the next component brought in for repair is an audio component, and let \(B\) be the event that the next component is a compact disc player (so the event \(B\) is contained in \(A\) ). Suppose that \(P(A)=.6\) and \(P(B)=.05\). What is \(P(B \mid A)\) ?

Short Answer

Expert verified
The conditional probability \(P(B \mid A)\) is approximately 0.0833.

Step by step solution

01

Understanding Conditional Probability

The problem asks for the conditional probability \(P(B \mid A)\), which is the probability that event \(B\) occurs given that event \(A\) has already occurred.
02

Applying the Conditional Probability Formula

We use the formula for conditional probability: \[ P(B \mid A) = \frac{P(B \cap A)}{P(A)} \] where \(P(B \cap A)\) is the probability that both events \(A\) and \(B\) occur.
03

Identifying Intersection

Since event \(B\) (compact disc player) is contained within event \(A\) (audio component), every occurrence of \(B\) is also an occurrence of \(A\). Therefore, \(P(B \cap A) = P(B) = 0.05\).
04

Calculating \(P(B \mid A)\)

Substitute the known values into the conditional probability formula: \[ P(B \mid A) = \frac{P(B \cap A)}{P(A)} = \frac{0.05}{0.6} \]
05

Performing the Division

Calculate \(\frac{0.05}{0.6}\) to find the conditional probability: \[ P(B \mid A) = \frac{0.05}{0.6} = 0.0833\]

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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 fundamental branch of mathematics that deals with understanding randomness and uncertainty. It assigns numerical values to events which measure the likelihood of these events occurring. These values range from 0 (meaning the event never occurs) to 1 (meaning the event always occurs).
In probability theory, we often deal with events and the relationships between them. A key aspect is calculating the probability of one event based on the knowledge of another event, known as conditional probability. Understanding this connection can help in practical scenarios such as predicting outcomes and making informed decisions.
In the provided exercise, conditional probability helps in determining the likelihood of a specific event (the next component being a compact disc player) under the condition that another event (it being an audio component) has occurred. This showcases how probability theory aids in breaking down and evaluating complex situations.
Intersection of Events
The intersection of events in probability is essentially about finding the probability that two events occur simultaneously. It is represented as the occurrence of both events, denoted as \(B \cap A\).
When you think of \(B \cap A\), you're considering situations where both event \(A\) (next component is an audio component) and event \(B\) (next component is a CD player) happen at the same time.
In the problem, since event \(B\) is a subset of event \(A\), this intersection \(P(B \cap A)\) simply equals \(P(B)\). This is because every time you have a CD player, it automatically means you have an audio component, making the two occurrences identical in this context.
  • This is an essential concept for solving problems where events are dependent or contained within each other.
Event Containment
Event Containment is a scenario in probability where one event is entirely within the scope of another event. If event \(B\) is contained within event \(A\), every occurrence of \(B\) is also an occurrence of \(A\).
In simpler terms, you can visualize event \(A\) as a larger circle with \(B\) as a smaller circle inside it. So, whenever B happens, A happens as well.
In the provided example, a compact disc player is always an audio component. Hence, the event \(B\) (next component is a CD player) is contained within event \(A\) (next component is an audio component). This relationship helps simplify the calculation of probabilities, since it eliminates uncertainties about their simultaneous occurrences.
Mathematical Statistics
Mathematical Statistics involves the use of probability theory to analyze and interpret data collected through experiments or surveys. It is crucial for making predictions and informed decisions based on data.
In the realm of probability, mathematical statistics helps in understanding the relationships and dependencies between events. For instance, calculating conditional probabilities like \(P(B \mid A)\) informs us about how data should be interpreted when one condition is assumed to be true.
  • In our problem scenario, mathematical statistics helps us deduce the likelihood of repairing a CD player, given that the device is an audio component. Such insights are invaluable for businesses aiming to optimize services and inventory based on probable requirements of customers.
By using the formula and given probabilities, mathematics supports practical and real-world problem-solving.

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Most popular questions from this chapter

Fifteen telephones have just been received at an authorized service center. Five of these telephones are cellular, five are cordless, and the other five are corded phones. Suppose that these components are randomly allocated the numbers 1 , \(2, \ldots, 15\) to establish the order in which they will be serviced. a. What is the probability that all the cordless phones are among the first ten to be serviced? b. What is the probability that after servicing ten of these phones, phones of only two of the three types remain to be serviced? c. What is the probability that two phones of each type are among the first six serviced?

Let \(A\) denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let \(B\) be the event that the next request is for help with SAS. Suppose that \(P(A)=.30\) and \(P(B)=.50\). a. Why is it not the case that \(P(A)+P(B)=1\) ? b. Calculate \(P\left(A^{\prime}\right)\). c. Calculate \(P(A \cup B)\). d. Calculate \(P\left(A^{\prime} \cap B^{\prime}\right)\).

An academic department has just completed voting by secret ballot for a department head. The ballot box contains four slips with votes for candidate \(A\) and three slips with votes for candidate \(B\). Suppose these slips are removed from the box one by one. a. List all possible outcomes. b. Suppose a running tally is kept as slips are removed. For what outcomes does \(A\) remain ahead of \(B\) throughout the tally?

Components of a certain type are shipped to a supplier in batches of ten. Suppose that \(50 \%\) of all such batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0,1 , and 2 defective components being in the batch under each of the following conditions? a. Neither tested component is defective. b. One of the two tested components is defective. [Hint: Draw a tree diagram with three firstgeneration branches for the three different types of batches.]

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost . Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Manag., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

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