Oh, An Average Box Containing 10 Articles Likely Has 2 Defective. If We Consider A Consignment Of 100 Boxes, How Many Of Them Are Expected To Have Three Or Fewer Defective?
Introduction
In this article, we will delve into the world of probability and statistics, specifically focusing on the binomial distribution. We will explore a classic problem that involves a consignment of boxes, each containing a certain number of articles. The problem states that an average box containing 10 articles likely has 2 defective. We are asked to determine how many of the 100 boxes in the consignment are expected to have three or fewer defective articles.
The Binomial Distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials, where each trial has a constant probability of success. In this case, we can consider each box as a trial, and the number of defective articles as the number of successes.
Defining the Parameters
Let's define the parameters of the binomial distribution:
- n: The number of trials (boxes) = 100
- p: The probability of success (defective article) = 2/10 = 0.2
- q: The probability of failure (non-defective article) = 1 - p = 0.8
Calculating the Probability of Three or Fewer Defective Articles
To calculate the probability of three or fewer defective articles in a box, we need to calculate the probability of 0, 1, 2, and 3 defective articles.
Calculating the Probability of 0 Defective Articles
The probability of 0 defective articles is given by the binomial distribution formula:
P(X = 0) = (nC0) * (p^0) * (q^(n-0))
where nC0 is the number of combinations of n items taken 0 at a time, which is equal to 1.
P(X = 0) = 1 * (0.2^0) * (0.8^100) = 0.0173
Calculating the Probability of 1 Defective Article
The probability of 1 defective article is given by the binomial distribution formula:
P(X = 1) = (nC1) * (p^1) * (q^(n-1))
where nC1 is the number of combinations of n items taken 1 at a time, which is equal to n.
P(X = 1) = 100 * (0.2^1) * (0.8^99) = 0.1349
Calculating the Probability of 2 Defective Articles
The probability of 2 defective articles is given by the binomial distribution formula:
P(X = 2) = (nC2) * (p^2) * (q^(n-2))
where nC2 is the number of combinations of n items taken 2 at a time, which is equal to n(n-1)/2.
P(X = 2) = 4950 * (0.2^2) * (0.8^98) = 0.2683
Calculating the Probability of 3 Defective Articles
The probability of 3 defective articles is given by the binomial distribution formula:
P(X = 3) = (nC3) * (p^3) * (q^(n-3))
nC3 is the number of combinations of n items taken 3 at a time, which is equal to n(n-1)(n-2)/6.
P(X = 3) = 161700 * (0.2^3) * (0.8^97) = 0.0421
Calculating the Total Probability of Three or Fewer Defective Articles
To calculate the total probability of three or fewer defective articles, we need to add the probabilities of 0, 1, 2, and 3 defective articles.
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0173 + 0.1349 + 0.2683 + 0.0421 = 0.4626
Conclusion
In this article, we used the binomial distribution to calculate the probability of three or fewer defective articles in a consignment of 100 boxes. We found that the probability of three or fewer defective articles is approximately 0.4626, or 46.26%. This means that we can expect approximately 46.26% of the boxes in the consignment to have three or fewer defective articles.
Implications
The results of this calculation have important implications for quality control and inventory management. By understanding the probability of defective articles in a consignment, manufacturers and suppliers can take steps to minimize the number of defective articles and improve the overall quality of their products.
Future Work
In future work, we could explore other applications of the binomial distribution, such as calculating the probability of a certain number of successes in a fixed number of trials. We could also investigate the use of other probability distributions, such as the Poisson distribution, to model the number of defective articles in a consignment.
References
- [1] Johnson, N. L., & Kotz, S. (1969). Distributions in statistics: discrete distributions. Wiley.
- [2] Feller, W. (1968). An introduction to probability theory and its applications. Wiley.
- [3] Ross, S. M. (2014). Introduction to probability models. Academic Press.
Q&A: Understanding the Binomial Distribution and Defective Articles ====================================================================
Introduction
In our previous article, we explored the binomial distribution and its application to a problem involving defective articles in a consignment of boxes. We calculated the probability of three or fewer defective articles in a box and found that the probability is approximately 0.4626, or 46.26%. In this article, we will answer some frequently asked questions (FAQs) related to the binomial distribution and defective articles.
Q: What is the binomial distribution?
A: The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials, where each trial has a constant probability of success.
Q: What are the parameters of the binomial distribution?
A: The parameters of the binomial distribution are:
- n: The number of trials (boxes) = 100
- p: The probability of success (defective article) = 2/10 = 0.2
- q: The probability of failure (non-defective article) = 1 - p = 0.8
Q: How do I calculate the probability of three or fewer defective articles?
A: To calculate the probability of three or fewer defective articles, you need to calculate the probability of 0, 1, 2, and 3 defective articles using the binomial distribution formula:
P(X = k) = (nCk) * (p^k) * (q^(n-k))
where nCk is the number of combinations of n items taken k at a time.
Q: What is the probability of 0 defective articles?
A: The probability of 0 defective articles is given by the binomial distribution formula:
P(X = 0) = (nC0) * (p^0) * (q^(n-0)) = 0.0173
Q: What is the probability of 1 defective article?
A: The probability of 1 defective article is given by the binomial distribution formula:
P(X = 1) = (nC1) * (p^1) * (q^(n-1)) = 0.1349
Q: What is the probability of 2 defective articles?
A: The probability of 2 defective articles is given by the binomial distribution formula:
P(X = 2) = (nC2) * (p^2) * (q^(n-2)) = 0.2683
Q: What is the probability of 3 defective articles?
A: The probability of 3 defective articles is given by the binomial distribution formula:
P(X = 3) = (nC3) * (p^3) * (q^(n-3)) = 0.0421
Q: How do I calculate the total probability of three or fewer defective articles?
A: To calculate the total probability of three or fewer defective articles, you need to add the probabilities of 0, 1, 2, and 3 defective articles:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0173 + 0.1349 + 0.2683 + 0.0421 = 0.4626
Q: What are the implications of this calculation?
A: The results of this calculation have important implications for quality control and inventory management. By understanding the probability of defective articles in a consignment, manufacturers and suppliers can take steps to minimize the number of defective articles and improve the overall quality of their products.
Q: What are some future directions for this research?
A: Some future directions for this research include:
- Exploring other applications of the binomial distribution, such as calculating the probability of a certain number of successes in a fixed number of trials.
- Investigating the use of other probability distributions, such as the Poisson distribution, to model the number of defective articles in a consignment.
Conclusion
In this article, we answered some frequently asked questions related to the binomial distribution and defective articles. We hope that this Q&A article has provided a helpful resource for those interested in understanding the binomial distribution and its application to quality control and inventory management.