/*! 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 17 Suppose that a computer manufact... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. You can represent possible outcomes of the selection process by pairs. For example, the pair (1,2) represents the selection of Boards 1 and 2 for inspection. a. List the 10 different possible outcomes. b. Suppose that Boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Let \(x=\) the number of defective boards observed among those inspected. Find the probability distribution of \(x\).

Short Answer

Expert verified
The probability distribution of the number of defective boards observed, \(x\), is as follows: \(x \ P(X=x)\) \(0 \ \frac{3}{10}\) \(1 \ \frac{6}{10}\) \(2 \ \frac{1}{10}\)

Step by step solution

01

List the possible outcomes for selecting two boards from the lot.

To list the possible outcomes for selecting two boards from the lot, we'll consider every unique pair of the 5 boards. The pairs can be: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), and (4,5)
02

Calculate the probabilities of observing 0, 1, or 2 defective boards.

Let \(x\) represent the number of defective boards observed among those inspected. Since there are only two defective boards (1 and 2) in the lot, we can have the following three possible outcomes: \(x = 0\): No defective boards are chosen. \(x = 1\): Exactly one defective board is chosen. \(x = 2\): Both defective boards are chosen. We will now calculate the probability for each of these outcomes.
03

Calculate the probability for \(x = 0\).

To calculate the probability of observing no defective boards \((x = 0)\), we need to determine the number of possible outcomes with no defective boards and divide it by the total number of outcomes. Out of the 10 outcomes, the following outcomes contain no defective boards: (3,4), (3,5), and (4,5). So, there are 3 outcomes with no defective boards. The probability of observing no defective boards is: \(P(x = 0) = \frac{\text{number of outcomes with 0 defective boards}}{\text{total number of outcomes}} = \frac{3}{10}\)
04

Calculate the probability for \(x = 1\).

To calculate the probability of observing exactly one defective board \((x = 1)\), we need to determine the number of possible outcomes with only one defective board and divide it by the total number of outcomes. Out of the 10 outcomes, the following outcomes contain exactly one defective board: (1,3), (1,4), (1,5), (2,3), (2,4), and (2,5). So, there are 6 outcomes with exactly one defective board. The probability of observing exactly one defective board is: \(P(x = 1) = \frac{\text{number of outcomes with 1 defective board}}{\text{total number of outcomes}} = \frac{6}{10}\)
05

Calculate the probability for \(x = 2\).

To calculate the probability of observing both defective boards \((x = 2)\), we need to determine the number of possible outcomes with two defective boards and divide it by the total number of outcomes. Out of the 10 outcomes, there is only one outcome that contains both defective boards, which is: (1,2) The probability of observing both defective boards is: \(P(x = 2) = \frac{\text{number of outcomes with 2 defective boards}}{\text{total number of outcomes}} = \frac{1}{10}\)
06

The probability distribution of \(x\).

After calculating the probabilities for each value of \(x\), we can represent the probability distribution of the number of defective boards observed. The probability distribution of \(x\) is given by: • \(P(x = 0) = \frac{3}{10}\) • \(P(x = 1) = \frac{6}{10}\) • \(P(x = 2) = \frac{1}{10}\) So, the probability distribution of the number of defective boards observed is: \(x \ P(X=x)\) \(0 \ \frac{3}{10}\) \(1 \ \frac{6}{10}\) \(2 \ \frac{1}{10}\)

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.

Statistical Outcomes
Understanding statistical outcomes is essential for analyzing probability events. When we refer to statistical outcomes, we're talking about all the possible results that can occur from a probabilistic event. In the context of our exercise, the statistical outcomes are the distinct pairs of computer boards that can be selected for inspection from a lot of five boards.

Each pair is a potential outcome that needs to be considered when calculating probabilities. In our example, there are 10 different possible statistical outcomes: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), and (4,5). To ensure we accurately understand the problem, it's vital to list all these outcomes clearly as it forms the basis for further calculations involving probability.
Defective Board Probability
The probability of picking a defective board is a practical problem that can occur in manufacturing processes. It involves analysis of random events to determine the likelihood of selecting a defective item from a batch. In our exercise, two boards out of five are defective, and we want to know the probability of picking none, one, or both of these defective boards when two are randomly selected for inspection.

To improve the student's comprehension of the concept, it's crucial to walk through the probability calculations for each scenario. For instance, we found the probability of selecting no defective boards (\(x = 0\)) to be \frac{3}{10}, since there are 3 outcomes without defective boards. Similarly, the calculation for selecting exactly one defective board (\(x = 1\)) yields a probability of \frac{6}{10}, and selecting both defective boards (\(x = 2\)) has a probability of \frac{1}{10}. This step-by-step approach aids in reinforcing the student's understanding of probabilities in discrete events.
Combinatorics
Combinatorics is the branch of mathematics dealing with the study of counting, combination, and permutation of sets. It's a powerful tool for calculating the number of possible outcomes in problems involving probability. We applied combinatorics in our exercise to determine the 10 possible outcomes for choosing pairs of boards.

In more complex problems, combinatorial formulas such as the factorial, permutations, and combinations come into play. These formulas are essential for calculating the number of ways that events can occur, which in turn helps us to compute probabilities. For example, in our exercise, we intuitively listed all pairs, but in larger sets, combinatorial formulas would be necessary to avoid errors and to efficiently calculate the total number of possible outcomes without listing them all.

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

Suppose that in a certain metropolitan area, \(90 \%\) of all households have cable TV. Let \(x\) denote the number among four randomly selected households that have cable TV. Then \(x\) is a binomial random variable with \(n=4\) and \(p=0.9\). a. Calculate \(p(2)=P(x=2),\) and interpret this probability. b. Calculate \(p(4),\) the probability that all four selected households have cable TV. c. Calculate \(P(x \leq 3)\).

Sophie is a dog who loves to play catch. Unfortunately, she isn't very good at this, and the probability that she catches a ball is only \(0.1 .\) Let \(x\) be the number of tosses required until Sophie catches a ball. a. Does \(x\) have a binomial or a geometric distribution? b. What is the probability that it will take exactly two tosses for Sophie to catch a ball? c. What is the probability that more than three tosses will be required?

A soft-drink machine dispenses only regular Coke and Diet Coke. Sixty percent of all purchases from this machine are diet drinks. The machine currently has 10 cans of each type. If 15 customers want to purchase drinks before the machine is restocked, what is the probability that each of the 15 is able to purchase the type of drink desired?

Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random from a lot and inspect them. Suppose that a lot is judged acceptable if one or fewer of these 20 parts are defective. If more than one part is defective, the lot is rejected and returned to the supplier. Find the probability of accepting lots that have each of the following (Hint: Identify success with a defective part.): a. \(5 \%\) defective parts b. \(10 \%\) defective parts c. \(20 \%\) defective parts

The paper referenced in Example 6.24 ("Estimating Waste Transfer Station Delays Using GPS," Waste Management [2008]: 1742-1750) describing processing times for garbage trucks also provided information on processing times at a second facility. At this second facility, the mean total processing time was 9.9 minutes and the standard deviation of the processing times was 6.2 minutes. Explain why a normal distribution with mean 9.9 and standard deviation 6.2 would not be an appropriate model for the probability distribution of the variable \(x=\) total processing time of a randomly selected truck entering this second facility.

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.