Probability Distribution of Defective Items in a Box

Understanding the probability distribution of defective items in a sample drawn from a larger population is a common problem in statistics. This article explores the concept in depth by examining a specific scenario involving a box containing 12 items, 3 of which are defective. This problem is solved using the Hypergeometric Distribution, which is suitable for situations without replacement.

Scenario and Probability Distribution

Consider a box containing 12 items, of which 3 are defective. When a sample of 3 items is randomly selected from this box, we aim to find the probability distribution of the number of defective items in this sample. Since the selection is without replacement, the Hypergeometric Distribution is the appropriate method to use.

Understanding the Hypergeometric Distribution

The Hypergeometric Distribution is used when we are performing a random sampling without replacement. It is defined by the probability mass function (PMF), which is given by:

[math]P(X k) frac{{M choose k}{N - M choose n - k}}{N choose n}

XX: The random variable representing the number of defective items in the sample. MM: The total number of defective items in the box (3 in this case). NN: The total number of items in the box (12 in this case). NN: The number of items in the sample (3 in this case). kk: The number of defective items we are interested in in the sample.

Calculating the Probability Distribution

Given that we want to find the probability of having 0, 1, 2, or 3 defective items in the sample, we will calculate the corresponding probabilities using the Hypergeometric Distribution formula.

Number of Defective Items: 0

The probability of selecting 0 defective items in the sample can be calculated as follows:

P(X 0) frac{{3 choose 0}{12 - 3 choose 3 - 0}}{12 choose 3} frac{1 times frac{9!}{3!6!}}{frac{12!}{3!9!}} frac{84}{220} frac{21}{55}

Number of Defective Items: 1

The probability of selecting exactly 1 defective item in the sample is given by:

P(X 1) frac{{3 choose 1}{12 - 3 choose 3 - 1}}{12 choose 3} frac{3 times frac{9!}{2!7!}}{frac{12!}{3!9!}} frac{3 times 36}{220} frac{108}{220} frac{27}{55}

Number of Defective Items: 2

The probability of selecting 2 defective items in the sample is:

P(X 2) frac{{3 choose 2}{12 - 3 choose 3 - 2}}{12 choose 3} frac{3 times frac{9!}{1!8!}}{frac{12!}{3!9!}} frac{3 times 9}{220} frac{27}{220}

Number of Defective Items: 3

The probability of selecting all 3 defective items in the sample is:

P(X 3) frac{{3 choose 3}{12 - 3 choose 3 - 3}}{12 choose 3} frac{1 times frac{9!}{0!9!}}{frac{12!}{3!9!}} frac{1 times 1}{220} frac{1}{220}

Expected Number of Defective Items

We can determine the expected number of defective items in the sample by calculating the expected value of the Hypergeometric Distribution:

E(X) frac{M times n}{N} frac{3 times 3}{12} frac{9}{12} frac{3}{4} 0.75

Thus, the expected number of defective items in a sample of 3 from a box containing 12 items, 3 of which are defective, is 0.75.

Conclusion

The Hypergeometric Distribution is a useful tool for understanding the probability distribution of defective items in a sample drawn from a larger population. In this article, we used it to calculate the probabilities and expected number of defective items in a scenario where a sample of 3 items is drawn from a box of 12 items, 3 of which are defective.

Understanding these concepts can be vital in a variety of practical applications, from manufacturing quality control to epidemiological studies.