nas, III past, its goblets. Such goblets are designated as "seconds". Assume defective goblets are independe a) With what probability will there be one "second" among six randomly selected goblets? b) with what probability will there be at least four "seconds" among twelve randomly selected goblets? c) What is the expected number of "seconds" among 50 goblets? What is the associated variance?
nas, III past, its goblets. Such goblets are designated as "seconds". Assume defective goblets are independe a) With what probability will there be one "second" among six randomly selected goblets? b) with what probability will there be at least four "seconds" among twelve randomly selected goblets? c) What is the expected number of "seconds" among 50 goblets? What is the associated variance?
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![### Crystal Goblet Production Defect Probability
**A company producing fine crystal goblets has, in the past, experienced a defect rate of 10% in its goblets. Such goblets are designated as “seconds”. Assume defective goblets are independent.**
**a) With what probability will there be one “second” among six randomly selected goblets?**
**b) With what probability will there be at least four “seconds” among twelve randomly selected goblets?**
**c) What is the expected number of “seconds” among 50 goblets? What is the associated variance?**
#### Hint: Binomial Distribution
---
This problem involves using the binomial distribution to determine the probability of different numbers of defective goblets ("seconds") in samples of various sizes. The binomial distribution is applicable here because we have a fixed number of trials (goblets selected), each trial results in either a success (defective goblet) or failure (non-defective goblet), and the probability of success is constant.
In this case, the probability of a goblet being defective (success) is 10%, or 0.10. Let’s break down each part of the question:
**a) Probability of One “Second” Among Six Goblets:**
This asks for the probability of exactly one defective goblet in a sample of six. The binomial probability formula is:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( n \) is the number of trials (goblets), \( n = 6 \)
- \( k \) is the number of successes (defective goblets), \( k = 1 \)
- \( p \) is the probability of success, \( p = 0.10 \)
**b) Probability of At Least Four “Seconds” Among Twelve Goblets:**
To find the probability of getting at least four defective goblets in a sample of twelve, you would calculate the cumulative probability for four through twelve defective goblets and sum these probabilities.
**c) Expected Number of “Seconds” and Variance Among 50 Goblets:**
The expected number (mean) of defective goblets in a sample can be found using:
\[ \mu = np \]
Where \( n \) is the number of trials and \( p \) is the probability of success.
The variance is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb19fb600-654f-4bb5-880f-15449a731741%2F95ae3497-da82-406a-be31-eff53c8bc946%2Fonq7xdg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Crystal Goblet Production Defect Probability
**A company producing fine crystal goblets has, in the past, experienced a defect rate of 10% in its goblets. Such goblets are designated as “seconds”. Assume defective goblets are independent.**
**a) With what probability will there be one “second” among six randomly selected goblets?**
**b) With what probability will there be at least four “seconds” among twelve randomly selected goblets?**
**c) What is the expected number of “seconds” among 50 goblets? What is the associated variance?**
#### Hint: Binomial Distribution
---
This problem involves using the binomial distribution to determine the probability of different numbers of defective goblets ("seconds") in samples of various sizes. The binomial distribution is applicable here because we have a fixed number of trials (goblets selected), each trial results in either a success (defective goblet) or failure (non-defective goblet), and the probability of success is constant.
In this case, the probability of a goblet being defective (success) is 10%, or 0.10. Let’s break down each part of the question:
**a) Probability of One “Second” Among Six Goblets:**
This asks for the probability of exactly one defective goblet in a sample of six. The binomial probability formula is:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Where:
- \( n \) is the number of trials (goblets), \( n = 6 \)
- \( k \) is the number of successes (defective goblets), \( k = 1 \)
- \( p \) is the probability of success, \( p = 0.10 \)
**b) Probability of At Least Four “Seconds” Among Twelve Goblets:**
To find the probability of getting at least four defective goblets in a sample of twelve, you would calculate the cumulative probability for four through twelve defective goblets and sum these probabilities.
**c) Expected Number of “Seconds” and Variance Among 50 Goblets:**
The expected number (mean) of defective goblets in a sample can be found using:
\[ \mu = np \]
Where \( n \) is the number of trials and \( p \) is the probability of success.
The variance is
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