/*! 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 193 Two assembly lines I and II have... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two assembly lines I and II have the same rate of defectives in their production of voltage regulators. Five regulators are sampled from each line and tested. Among the total of ten tested regulators, four are defective. Find the probability that exactly two of the defective regulators came from line I.

Short Answer

Expert verified
The probability is approximately 0.476.

Step by step solution

01

Define Random Variables

Let the random variable \( X \) represent the number of defective regulators coming from line I. The distribution of \( X \) is a hypergeometric distribution since we are drawing without replacement from two different samples.
02

Identify Parameters for Hypergeometric Distribution

We have a total of \( N = 10 \) regulators, with \( K = 4 \) defectives among them; and we choose \( n = 5 \) regulators from line I. We want to find the probability that exactly \( k = 2 \) defectives are from line I.
03

Formula for Hypergeometric Probability

The probability is given by the hypergeometric distribution formula: \[ P(X = k) = \frac{{\binom{K}{k} \cdot \binom{N-K}{n-k}}}{{\binom{N}{n}}} \] Substitute the respective values into the formula.
04

Compute Combinatorial Values

Calculate each term:- \( \binom{4}{2} = 6 \)- \( \binom{6}{3} = 20 \)- \( \binom{10}{5} = 252 \).
05

Compute Probability

Substitute the calculated combinatorial values into the formula:\[ P(X = 2) = \frac{6 \times 20}{252} = \frac{120}{252} \approx 0.4762 \].
06

Interpret Result

The probability that exactly two of the defective regulators came from line I is approximately 0.4762.

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.

Combinatorial Probability
Combinatorial probability is a branch of probability theory that involves counting and arranging different outcomes. In simple words, it deals with how many ways certain events can happen. This concept often employs combinations, which are ways to select items from a group without considering the order. The main formula for combinations is given by:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \( n \) is the total number of items, \( k \) is the number of items to choose, and \(!\) denotes a factorial, which is the product of all positive integers up to a certain number.
Combinations are essential in calculating probabilities in scenarios where order doesn’t matter. For instance, in the hypergeometric distribution, we use combinatorial expressions to determine the probability of a certain number of successes in a sample, as shown in the exercise above.
Defective Items
Defective items refer to products that fail to meet the quality standards set by a manufacturer. These can occur due to various factors like faulty materials or errors in the manufacturing process. In terms of probability, defective items introduce uncertainty, particularly in quality control and sampling. Companies often take samples from their production lines to estimate the defective rate. It's a practical approach because inspecting each item individually is generally impractical. In this exercise, we have ten voltage regulators, of which four are known to be defective. By calculating the probability of picking defective items from a sample, companies can make informed decisions regarding the quality of their products.
Sampling Without Replacement
Sampling without replacement is a method where once an item is selected from a group, it is not returned to the group for subsequent selections. This results in reducing the total number of items available for the next draw. This concept greatly affects the calculation of probability because the probability of drawing an item changes with each selection. Consequently, it impacts distributions like the hypergeometric distribution, where probabilities are calculated based on the number of successful outcomes in samples where items are not replaced.
In the given exercise, regulators are sampled without replacement, which is why the hypergeometric distribution is applicable. This method reflects real-world situations more accurately, as it's a common practice in quality testing and lotteries.
Random Variables
Random variables are fundamental to probability theory. A random variable is essentially a function that assigns numerical values to outcomes of a random phenomenon. There are two main types of random variables:
  • Discrete random variables that assume a countable number of possible values. For example, the number of heads in coin tosses.
  • Continuous random variables, which take values within a continuous range, like the height of people.
In this context, the random variable \( X \), represents the number of defective regulators from line I. Since it deals with a countable outcome, with multiple possibilities (0 to 5 defectives from line I), \( X \) is a discrete random variable. This concept helps us define and compute the probabilities associated with different outcomes using probability distributions like the hypergeometric distribution.

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 \(Y\) is a discrete random variable with mean \(\mu\) and variance \(\sigma^{2}\) and let \(W=2 Y\) a. Do you expect the mean of \(W\) to be larger than, smaller than, or equal to \(\mu=E(Y) ?\) Why? b. Use Theorem 3.4 to express \(E(W)=E(2 Y)\) in terms of \(\mu=E(Y) .\) Does this result agree with your answer to part (a)? c. Recalling that the variance is a measure of spread or dispersion, do you expect the variance of \(W\) to be larger than, smaller than, or equal to \(\sigma^{2}=V(Y) ?\) Why? d. Use Definition 3.5 and the result in part (b) to show that $$V(W)=E\left\\{[W-E(W)]^{2}\right\\}=E\left[4(Y \mu)^{2}\right]=4 \sigma^{2}$$; that is, \(W=2 Y\) has variance four times that of \(Y\).

The number of knots in a particular type of wood has a Poisson distribution with an average of 1.5 knots in 10 cubic feet of the wood. Find the probability that a 10 -cubic-foot block of the wood has at most 1 knot.

A multiple-choice examination has 15 questions, each with five possible answers, only one of which is correct. Suppose that one of the students who takes the examination answers each of the questions with an independent random guess. What is the probability that he answers at least ten questions correctly?

Let \(Y\) be a random variable with mean 11 and variance \(9 .\) Using Tchebysheff's theorem, find a. a lower bound for \(P(6

In the 18th century, the Chevalier de Mere asked Blaise Pascal to compare the probabilities of two events. Below, you will compute the probability of the two events that, prior to contrary gambling experience, were thought by de Mere to be equally likely. a. What is the probability of obtaining at least one 6 in four rolls of a fair die? Answer \(t\) b. If a pair of fair dice is tossed 24 times, what is the probability of at least one double six?

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.