/*! 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 95 Components of a certain type are... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 95

Components of a certain type are shipped to a supplier in batches of \(10 .\) Suppose that \(50 \%\) of all batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. A batch is selected at random. Two components from this batch are randomly selected and tested. a. If the batch from which the components were selected actually contains two defective components, what is the probability that neither of these is selected for testing? b. What is the probability that the batch contains two defective components and that neither of these is selected for testing? c. What is the probability that neither component selected for testing is defective? (Hint: This could happen with any one of the three types of batches. A tree diagram might help.)

Short Answer

Expert verified
a. The probability neither component is defective if there are two defective ones in the batch is \( 28/45 \), \n b. The probability that the batch contains two defective components and neither is defective is \(0.2 * 28/45\), \n c. The probability that neither component selected for testing is defective is \(0.5*1 + 0.3*(C(9,2)/C(10,2)) + 0.2*(C(8,2)/C(10,2))\).

Step by step solution

01

Probability neither component is defective if there are two in the batch

We determine the probability neither component is defective if there are two defective ones in the batch using combinations. We have two defective components, meaning there are 8 working ones. Choosing 2 out of those 8 without replacement gives us \(C(8,2) = 28\) ways. There are total of \(C(10,2) = 45\) ways we can choose 2 components from the batch. Hence the probability will be the ratio of those two, which equates to \(P = 28/45\).
02

Probability batch contains two defective components but neither is chosen

To determine the probability that the batch contains two defective components and neither is tested, we must consider the probability of selecting a batch that contains exactly two defective components. We know that this is 20 percent, or 0.2. Hence, the desired probability is product of the probability of selecting a batch that contain two defective components and neither is chosen for testing, which is \(0.2 * (28/45)\).
03

Calculate probability that neither component selected for testing is defective

Here we have three types of batches, one with 0 defective components, one with 1 defective component, and lastly one with two defective components. The probabilities of choosing a non-defective component from each of these is respectively \(1, C(9,2) / C(10,2), C(8,2) / C(10,2)\), and the probabilities of these batches are 0.5, 0.3, and 0.2. So, the total probability will be a sum of each of these probabilities multiplied by corresponding batch probability. That gives us the result: \(0.5*1 + 0.3*(C(9,2)/C(10,2)) + 0.2*(C(8,2)/C(10,2))\).

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 91影视!

Key Concepts

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

Probability with Defective Components
Understanding the probability of encountering defective components in a batch is crucial in quality control and risk management. Let's imagine a scenario where you have multiple batches of products, and some contain defective items. The probability of selecting a defective or non-defective component depends on the composition of the batch you're dealing with.

For instance, if you know a certain batch contains two defective components out of ten, you could ask: What are the odds of randomly picking two items and both being non-defective? This problem requires combinatorial probability knowledge to solve, which involves calculating the number of favorable outcomes over all possible outcomes. Specifically, if you have eight non-defective components, there are C(8,2) = 28 ways to choose two non-defective components out of the possible C(10,2) = 45 ways to choose any two components.

With these combinations, you can compute the probability by dividing the number of favorable outcomes (selecting non-defective components) by the total number of possible outcomes (selecting any two components). So, if you aim to improve your proficiency in dealing with defective components in statistics, grasping the foundations of combinatorial calculations and thinking critically about batch compositions are essential steps.
Combinatorial Probability
Combinatorial probability is a field of mathematics that deals with counting and calculating the likelihood of events occurring, based on the combination of various outcomes. In our example, we use combinations to determine the number of ways we can select two non-defective components from a batch.

Combinations are a specific way to calculate how many different groups can be formed from a larger set of items, where the order of the items does not matter. The formula for calculating the combination of choosing k items from a set of n is defined as C(n,k) = n! / (k!(n-k)!), where ! denotes factorial, the product of all positive integers up to that number.

Real-World Applications

Combinatorial probability isn't just theoretical; it finds applications in various fields such as genetics, marketing, finance, and more. For instance, it helps geneticists calculate the odds of inheriting specific traits, while marketers may use it to predict the potential combinations consumers might choose when selecting product features.
Tree Diagram in Probability
A tree diagram is a visual representation that helps to map out all possible outcomes of a probability event. It's like a branching diagram that shows all possible paths from start to finish, where each branch represents a possible decision or chance event.

Using a tree diagram is especially useful when dealing with multiple stages of probability events. In our textbook exercise, for instance, to calculate the probability that neither component selected for testing is defective, we can use a tree diagram to visualize the three different types of batches (with 0, 1, or 2 defective components).

Each branch of the tree diagram would represent the probabilities of selecting each batch type, and further branches would represent the probabilities of selecting non-defective components from those batches. By multiplying along the branches and adding the probabilities of each 'end' branch, we can determine the overall probability of selecting non-defective components. Tree diagrams provide an intuitive way to understand complex probability questions and can be an invaluable tool for students to grasp multi-stage probability events.

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

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7, P(F)=.6\), and \(P(E \cap F)=.54\). a. Calculate \(P(E \mid F)\) the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

An article in the New York Times (March 2, 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P\) (survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

Refer to the following information on births in the United States over a given period of time: $$ \begin{array}{lr} \text { Type of Birth } & \text { Number of Births } \\ \hline \text { Single birth } & 41,500,000 \\ \text { Twins } & 500,000 \\ \text { Triplets } & 5000 \\ \text { Quadruplets } & 100 \\ & \\ \hline \end{array} $$ Use this information to approximate the probability that a randomly selected pregnant woman who reaches full term a. Delivers twins b. Delivers quadruplets c. Gives birth to more than a single child

Many fire stations handle emergency calls for medical assistance as well as calls requesting firefighting equipment. A particular station says that the probability that an incoming call is for medical assistance is \(.85 .\) This can be expressed as \(P(\) call is for medical assistance \()=.85\). a. Give a relative frequency interpretation of the given probability. b. What is the probability that a call is not for medical assistance? c. Assuming that successive calls are independent of one another, calculate the probability that two successive calls will both be for medical assistance. d. Still assuming independence, calculate the prohahility that for two successive calls, the first is for medical assistance and the second is not for medical assistance. e. Still assuming independence, calculate the probability that exactly one of the next two calls will be for medical assistance. (Hint: There are two different possibilities. The one call for medical assistance might be the first call, or it might be the second call.) f. Do you think that it is reasonable to assume that the requests made in successive calls are independent? Explain.

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a schedule is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is \(.8\). 2\. If Maria completes her part on time, the probability that Alex completes on time is \(.9\), but if Maria is late, the probability that Alex completes on time is only . 6 . 3\. If Alex completes his part on time, the probability that Juan completes on time is \(.8\), but if Alex is late, the probability that Juan completes on time is only .5. 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(.9\), but if Juan is late, the probability that Jacob completes on time is only \(.7 .\) Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation
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.