/*! 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 52 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
\(P(B \mid A) \approx 0.0833\).

Step by step solution

01

Understanding the Conditional Probability

The problem asks for the probability of an event occurring under a condition. Here, we want to find the probability that the next component is a compact disc player (event \(B\)) given that it is an audio component (event \(A\)). This is denoted as \(P(B \mid A)\).
02

Using the Formula for Conditional Probability

The formula for conditional probability is \(P(B \mid A) = \frac{P(B \cap A)}{P(A)}\). First, recognize that since \(B\) is contained within \(A\), \(P(B \cap A) = P(B)\). Therefore, we can substitute \(P(B \cap A)\) with \(P(B)\) in this formula.
03

Substitute Known Values

Substitute the known values into the formula. We know \(P(B) = 0.05\) and \(P(A) = 0.6\). So our equation becomes: \[P(B \mid A) = \frac{0.05}{0.6}\].
04

Calculate the Conditional Probability

Perform the division to calculate \(P(B \mid A)\): \[P(B \mid A) = \frac{0.05}{0.6} \approx 0.0833\]. This is the probability that the next component is a compact disc player, given it is an audio component.

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Key Concepts

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

Sample Space
In probability, the concept of a sample space is foundational. Think of it as the universe of all possible outcomes for any given experiment or event. For example, if you’re tossing a coin, the sample space is {Heads, Tails}. It encompasses every potential result you might observe. In our exercise about repairing shop components, the sample space includes all possible types of components, both audio and video, that could be brought into the shop. It visually represents every potential scenario the shop may face regarding incoming repairs.
  • A sample space gives a complete set of all things that could happen.
  • Each outcome in the sample space is unique and exclusive.
Understanding the sample space helps us recognize the events, such as components being audio or video, within the complete range of possibilities. This recognition is crucial when we dive into specific probabilities tied to these events.
Event Probability
Event probability measures the likelihood of a certain outcome within a sample space. For instance, when we talk about something happening, like it raining tomorrow or pulling a red card from a deck, we're discussing event probability. In the context of our exercise, the event probability helps determine how likely it is for a component brought in for repair to be audio, denoted as event \(A\), or a compact disc player, denoted as event \(B\). Given the probabilities, \(P(A) = 0.6\) and \(P(B) = 0.05\), we can make informed predictions about these occurrences.
  • Event probability ranges from 0 to 1, where 0 means impossible and 1 means certain.
  • Having the probability of each event helps in evaluating how often we can expect it to occur.
By knowing the probabilities of events \(A\) and \(B\), we can calculate their interaction, such as the probability of \(B\) occurring given that \(A\) has occurred.
Intersection of Events
The intersection of events considers how two events might overlap or occur together within a sample space. We denote this as \(A \cap B\), meaning both event \(A\) and event \(B\) happen at the same time. In the exercise, we see that event \(B\) is actually contained within event \(A\), meaning every time an event \(B\) occurs, event \(A\) happens as well. Thus, \(P(B \cap A)\) is the same as \(P(B)\). Such intersections help in calculating conditional probability using the formula \(P(B \mid A) = \frac{P(B \cap A)}{P(A)}\).
  • An intersection narrows down the sample space to outcomes shared by both events.
  • The overlap represents joint occurrences within the broader sample space.
Understanding these intersections makes it easier to assess combined probabilities and predict scenarios where multiple conditions hold true simultaneously.

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Most popular questions from this chapter

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2,000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\) [the first board is green] and \(B=\) [the second board is green ). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think that an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

A mutual fund company offers its customers several different funds: a money- market fund, three different bond funds (short, intermediate, and long-term), two stock funds (moderate and high-risk), and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: \(\begin{array}{lrcr}\text { Money-market } & 20 \% & \text { High-risk stock } & 18 \% \\ \text { Short bond } & 15 \% & \text { Moderate-risk } & \\ \text { Intermediate } & & \text { stock } & 25 \% \\ \text { bond } & 10 \% & \text { Balanced } & 7 \% \\ \text { Long bond } & 5 \% & & \end{array}\) A customer who owns shares in just one fund is randomly selected. a. What is the probability that the selected individual owns shares in the balanced fund? b. What is the probability that the individual owns shares in a bond fund? c. What is the probability that the selected individual does not own shares in a stock fund?

l and #2. If one pump fails, the system will still operate. However, because of the added strain, the extra remaining pump is … # A system consists of two identical pumps, #l and #2. If one pump fails, the system will still operate. However, because of the added strain, the extra remaining pump is now more likely to fail than was originally the case. That is, \(r=P(\\# 2\) fails # I fails) \(>P(\) #2 fails \()=q\). If at least one pump fails by the end of the pump design life in \(7 \%\) of all systems and both pumps fail during that period in only \(1 \%\), what is the probability that pump #1 will fail during the pump design life?

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45). a. How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift? b. What is the probability that all 6 selected workers will be from the same shift? c. What is the probability that at least two different shifts will be represented among the selected workers? d. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

Use Venn diagrams to verify the following two relationships for any events \(A\) and \(B\) (these are called De Morgan's laws): a. \((A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}\) b. \((A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}\)

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