In manufacturing and production processes, it is crucial to understand the likelihood of non-defective items in a given sample. This can help in quality control and decision-making. Let's explore how to calculate the probability of non-defective items in a sample using both binomial and hypergeometric distributions.

Introduction

When a manufacturer wants to ensure the quality of their products, one of the key metrics is the probability of an item being defective. This information can be valuable for both internal quality control and external buyers. In this article, we will explore how to calculate the probability that none of the items in a sample is defective.

Sample Scenario: Defective Items in a Batch

Consider a scenario where 3 out of every 100 items in a batch are defective. This means the probability of an item being defective is 0.03, and the probability of an item not being defective is 0.97. The question arises: what is the probability that in a random sample of 4 items, none will be defective?

Using Binomial Distribution

The binomial distribution model is often used for situations where there are a fixed number of trials (in this case, 4 items), each with the same probability of success (in this case, the item not being defective) denoted as ( p ), and we want to find the probability of ( k ) successes (in this case, ( k 0 )).

The formula for the probability of exactly ( k ) successes in ( n ) trials is:

[ P(X k) binom{n}{k} p^k (1-p)^{n-k} ]

Substituting ( n 4 ), ( k 0 ), and ( p 0.97 ), we get:

[ P(0 text{ defective items}) binom{4}{0} (0.97)^4 (0.03)^0 0.97^4 0.88529281 ]

This means the probability that in a sample of 4 items none will be defective is approximately 88.5%.

Another Scenario: Faulty Cell Phones

A factory manufactures cell phones with a defect rate of 4%. If we randomly choose 5 phones, what is the probability that none of them are defective?

Using Binomial Distribution

The probability of a phone not being defective is 0.96. Using the binomial distribution formula again, we can calculate:

[ P(0 text{ defective phones}) binom{5}{0} (0.96)^5 (0.04)^0 0.96^5 0.8153727 ]

This means the probability that none of the 5 phones are defective is approximately 81.5%.

Calculating with Hypergeometric Distribution

In some cases, the lot size is relatively small, and the number of items sampled is a significant portion of the lot. In such situations, the binomial distribution may not be the most accurate model. Instead, the hypergeometric distribution is more appropriate. The hypergeometric distribution is used to calculate the probability of ( k ) successes (in this case, non-defective items) in ( n ) draws (in this case, 4 items) from a finite population (the lot) without replacement.

The formula for the hypergeometric distribution is:

[ P(X k) frac{binom{D}{k} binom{N-D}{n-k}}{binom{N}{n}} ]

Where ( D ) is the number of defective items in the lot, ( N ) is the total number of items in the lot, ( n ) is the number of items sampled, and ( k ) is the number of non-defective items in the sample.

For example, if the lot contains 100 items, of which 3 are defective, we want to find the probability that in a sample of 4 items, none will be defective:

[ P(X 0) frac{binom{3}{0} binom{97}{4}}{binom{100}{4}} frac{7144}{8085} approx 0.883612 ]

This means the probability that none of the 4 items in the sample are defective is approximately 88.4%.

Conclusion

The probability of non-defective items in a sample can be calculated using both binomial and hypergeometric distributions, depending on the size of the lot and the nature of the sampling. While the binomial distribution is simpler and more straightforward, the hypergeometric distribution is more accurate when sampling without replacement from a small lot. Understanding these probabilities is essential for maintaining quality control in manufacturing and ensuring customer satisfaction.