/*! 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 83 A procedure often used to contro... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A procedure often used to control the quality of name-brand food products utilizes a panel of five "tasters." Each member of the panel tastes three samples, two of which are from batches of the product known to have the desired taste and the other from the latest batch. Each taster selects the sample that is different from the other two. Assume that the latest batch does have the desired taste, and that there is no communication between the tasters. a. If the latest batch tastes the same as the other two batches, what is the probability that the taster picks it as the one that is different? b. What is the probability that exactly one of the tasters picks the latest batch as different? c. What is the probability that at least one of the tasters picks the latest batch as different?

Short Answer

Expert verified
Answer: The probability that at least one of the tasters picks the latest batch as different is 211/243.

Step by step solution

01

Understand the problem

In this case, there are three samples - two with the desired taste and one from the latest batch (which also has the desired taste). As a taster, there are three possibilities for choosing the odd one out: randomly picking the sample from the latest batch as different, or selecting one of the two desired taste samples.
02

Calculate probability of picking latest batch

As the latest batch tastes the same as the other two batches, the chances of a taster selecting it as different is purely random. Since there are three samples, the probability of picking the latest batch as different would be 1/3. #b. Probability that exactly one of the tasters selects the latest batch#
03

Analyze the given information

We have a panel of 5 tasters, and we want to know the probability that exactly one of them chooses the latest batch as different.
04

Use binomial probability

As the outcome for each taster is either picking the latest batch or not, there is a binomial distribution in this case with n=5 (tasters) and p=1/3 (probability of picking the latest batch as different). We will use the formula: P(X=k) = (nCk) * (p^k) * (1-p)^(n-k) Where X is the number of times the latest batch is chosen as different, k is the desired number of successful outcomes (in this case, only 1 taster choosing the latest batch), n is the number of trials (5 tasters), and p is the probability of a successful outcome.
05

Calculate the probability

Using the binomial probability formula: P(X=1) = (5C1) * (1/3)^1 * (2/3)^4 = 10 * 1/3 * 16/81 = 160/243
06

Part b: Answer

The probability that exactly one of the tasters picks the latest batch as different is 160/243. #c. Probability that at least one of the tasters selects the latest batch#
07

Analyze the problem

In this case, we want to find the probability that at least one of the tasters picks the latest batch, meaning, it might be one, two, three, four, or all five tasters.
08

Use complementary probability

We can solve this problem easily by finding the complementary probability, which is the probability of none of the tasters selecting the latest batch as different.
09

Calculate complementary probability

Using the binomial probability formula for the case when none of the tasters selects the latest batch: P(X=0) = (5C0) * (1/3)^0 * (2/3)^5 = 1 * 1 * 32/243 = 32/243 As the complementary probability is P(X=0), the probability of at least one taster selecting the latest batch as different is: P(X≥1) = 1 - P(X=0) = 1 - 32/243 = 211/243
10

Part c: Answer

The probability that at least one of the tasters picks the latest batch as different is 211/243.

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.

Binomial Distribution
The binomial distribution is a foundational concept in probability and statistics, frequently used when there are two possible outcomes for each trial, like success or failure. In the original exercise, the task was to determine the probability of tasters picking a particular sample as different from the others. Here, success is defined as a taster selecting the latest batch as different, while failure is not selecting it.

Given that a panel of five tasters is involved, and each has a 1/3 chance of picking the latest batch as the "odd one out," the scenario can be described using a binomial distribution with:
  • **n** = the number of tasters, hence **n = 5**.
  • **p** = probability of success (choosing the latest batch), so **p = 1/3**.
This distribution helps us calculate probabilities for different numbers of tasters selecting the latest batch as different. For example, calculating exactly one taster selecting it involves using the formula: \[ P(X=k) = \binom{n}{k} \times p^k \times (1-p)^{n-k} \]where \( X \) is the number of tasters choosing the latest batch, \( k \) is the number of successes desired, and \( \binom{n}{k} \) is the number of ways to choose \( k \) successes out of \( n \) trials.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about the properties of a population. It involves the concept of null and alternative hypotheses. In the context of the exercise, we can form a hypothesis regarding the tasting panel:
  • **Null Hypothesis (H0):** The latest batch tastes the same as the other two samples.
  • **Alternative Hypothesis (H1):** The latest batch is different in taste.
Although in the exercise, it's assumed that the batch tastes the same, testing if one or more tasters select the batch as different could be viewed as an exploration of deviation from the expectation.

If more tasters consistently identify the latest batch as different, we might reject the null hypothesis, suggesting that there may indeed be a taste difference that needs evaluation. However, since we're examining only the probability of random selection, strong conclusions would need more rigorous testing with repeated experiments to confirm any hypotheses about quality control.
Quality Control
Quality control is a critical aspect in production, ensuring products meet certain standards. The tasting procedure used in the exercise serves as an informal quality control method for name-brand foods. With five trained tasters, the aim is to ensure that each batch aligns with the expected quality levels.

The practice of using a tasting panel can help spot inconsistencies. Here are a few key goals of quality control:
  • Ensuring products meet specified quality requirements and customer expectations.
  • Identifying deviations in production that might affect taste or safety.
  • Applying systematic procedures, like sampling and testing, to maintain quality levels.
In this scenario, quality control would likely entail repeated testing across different batches, leveraging the probability calculations from hypothesis testing and binomial distribution to determine whether observed differences are random or signify a meaningful issue in production batches. Using statistical methods allows for informed decision-making that can significantly improve quality outcomes.

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

If a person is given the choice of an integer from 0 to 9 , is it more likely that he or she will choose an integer near the middle of the sequence than one at either end? a. If the integers are equally likely to be chosen, find the probability distribution for \(x\), the number chosen. b. What is the probability that a person will choose a \(4,5,\) or \(6 ?\) c. What is the probability that a person will not choose a \(4,5,\) or \(6 ?\)

The increased number of small commuter planes in major airports has heightened concern over air safety. An eastern airport has recorded a monthly average of five near-misses on landings and takeoffs in the past 5 years. a. Find the probability that during a given month there are no near-misses on landings and takeoffs at the airport. b. Find the probability that during a given month there are five near-misses c. Find the probability that there are at least five nearmisses during a particular month.

Insulin-dependent diabetes (IDD) is a common chronic disorder of children. This disease occurs most frequently in persons of northern European descent, but the incidence ranges from a low of \(1-2\) cases per 100,000 per year to a high of more than 40 per 100,000 in parts of Finland. \({ }^{11}\) Let us assume that an area in Europe has an incidence of 5 cases per 100,000 per year. a. Can the distribution of the number of cases of IDD in this area be approximated by a Poisson distribution? If so, what is the mean? b. What is the probability that the number of cases of IDD in this area is less than or equal to 3 per \(100,000 ?\) c. What is the probability that the number of cases is greater than or equal to 3 but less than or equal to 7 per \(100,000 ?\) d. Would you expect to observe 10 or more cases of IDD per 100,000 in this area in a given year? Why or why not?

\(P(x \leq k)\) in each case: a. \(n=20, p=.05, k=2\) b. \(n=15, p=.7, k=8\) c. \(n=10, p=.9, k=9\)

Parents who are concerned that their children are "accident prone" can be reassured, according to a study conducted by the Department of Pediatrics at the University of California, San Francisco. Children who are injured two or more times tend to sustain these injuries during a relatively limited time, usually 1 year or less. If the average number of injuries per year for school- age children is two, what are the probabilities of these events? a. A child will sustain two injuries during the year. b. A child will sustain two or more injuries during the year. c. A child will sustain at most one injury during the year.

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.