15.3 Examples of Hypergeometric Distribution

The Hypergeometric Distribution is a probability distribution that describes the probability of obtaining a specific number of successes in a sample drawn without replacement from a finite population. This means that once an item is selected, it is not returned to the population before the next selection.

The formula for the Hypergeometric Probability is:

$$P(X = x) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}$$

Where:

• $N$: The total size of the population.
• $k$: The number of success states in the population.
• $n$: The number of draws (i.e., quantity drawn in each trial).
• $x$: The number of observed successes in the sample.
• $\binom{a}{b}$: The binomial coefficient, read as "a choose b", calculated as $\frac{a!}{b!(a-b)!}$.

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Example 1: Employee Selection for Training

Scenario: A company has 120 employees, and 30 of them are certified in a specific skill. The management wants to randomly select 6 employees for a special training program. What is the probability that exactly 4 of the selected employees are certified?

Given:

• Total employees ($N$): 120
• Certified employees ($k$): 30
• Number of employees selected ($n$): 6
• Desired number of certified employees in the selection ($x$): 4

Applying the Hypergeometric Probability Formula:

$$P(X = x) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}$$

Substitute the given values:

$$P(X = 4) = \frac{\binom{30}{4} \binom{120-30}{6-4}}{\binom{120}{6}} = \frac{\binom{30}{4} \binom{90}{2}}{\binom{120}{6}}$$

Calculation:

• $\binom{30}{4} = \frac{30!}{4!(30-4)!} = \frac{30!}{4!26!} = 27,405$
• $\binom{90}{2} = \frac{90!}{2!(90-2)!} = \frac{90!}{2!88!} = 4,005$
• $\binom{120}{6} = \frac{120!}{6!(120-6)!} = \frac{120!}{6!114!} = 1,691,030,320$

Now, plug these values back into the formula:

$$P(X = 4) = \frac{27,405 \times 4,005}{1,691,030,320} = \frac{109,837,025}{1,691,030,320} \approx 0.065$$

Result: The probability that exactly 4 of the 6 selected employees are certified is approximately 0.065.

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Example 2: Product Quality Check

Scenario: A warehouse contains 40 electronic devices, and 6 of them are known to be defective. A quality control inspector randomly selects 10 devices for testing. What is the probability that exactly 2 of the selected devices are defective?

Given:

• Total devices ($N$): 40
• Defective devices ($k$): 6
• Sample size ($n$): 10
• Desired number of defective devices in the sample ($x$): 2

Applying the Hypergeometric Formula:

$$P(X = x) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}$$

Substitute the given values:

$$P(X = 2) = \frac{\binom{6}{2} \binom{40-6}{10-2}}{\binom{40}{10}} = \frac{\binom{6}{2} \binom{34}{8}}{\binom{40}{10}}$$

Calculation:

• $\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = 15$
• $\binom{34}{8} = \frac{34!}{8!(34-8)!} = \frac{34!}{8!26!} = 1,235,346$
• $\binom{40}{10} = \frac{40!}{10!(40-10)!} = \frac{40!}{10!30!} = 847,660,528$

Now, plug these values back into the formula:

$$P(X = 2) = \frac{15 \times 1,235,346}{847,660,528} = \frac{18,530,190}{847,660,528} \approx 0.0219$$

Result: The probability that exactly 2 of the 10 selected devices are defective is approximately 0.022.

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Conclusion

These examples illustrate the practical application of the Hypergeometric Distribution in real-world scenarios involving sampling without replacement. It is a valuable tool in fields such as:

• Quality Assurance: Assessing the probability of finding a certain number of defective items in a batch.
• Human Resources: Determining the likelihood of selecting a specific number of employees with certain qualifications for training or promotions.
• Lottery Systems: Calculating the odds of winning based on the number of matching balls drawn.
• Survey Sampling: Estimating population characteristics when samples are drawn without replacement.

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Interview Questions

1. What is the Hypergeometric Distribution and how is it applied in employee selection? The Hypergeometric Distribution models the probability of a specific number of successes in a sample drawn without replacement from a finite population. In employee selection, it can be used to find the probability of selecting a certain number of employees with a specific skill or characteristic from a larger group.

2. How do you calculate the probability of selecting a fixed number of certified employees? You use the Hypergeometric Probability formula, specifying the total number of employees, the number of certified employees, the size of the selected group, and the desired number of certified employees in that group.

3. Explain the formula for the Hypergeometric Probability and its components. The formula $P(X = x) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}$ calculates the probability. $\binom{k}{x}$ represents the number of ways to choose $x$ successes from $k$ available successes. $\binom{N-k}{n-x}$ represents the number of ways to choose the remaining $n-x$ items from the $N-k$ non-successes. $\binom{N}{n}$ represents the total number of ways to choose any $n$ items from the population $N$.

4. How does the Hypergeometric Distribution model quality control in manufacturing? It's used to determine the probability that a sample taken from a production lot contains a specific number of defective items, given the total number of items and the total number of defectives in the lot. This helps in deciding whether to accept or reject a batch.

5. What does “sampling without replacement” mean in these examples? It means that once an item (employee or device) is selected for the sample, it is not put back into the population before the next selection. Each selection reduces the size of the remaining population and potentially the number of available success/failure items.

6. Why are combinations used in the Hypergeometric Distribution formula? Combinations are used because the order in which items are selected does not matter. We are interested in the group of selected items, not the sequence in which they were drawn.

7. How do you compute the probability of defective products in a random sample? By applying the Hypergeometric Distribution formula, where $N$ is the total number of products, $k$ is the total number of defective products, $n$ is the sample size, and $x$ is the desired number of defective products in the sample.

8. What are some real-world applications of the Hypergeometric Distribution? Beyond quality control and employee selection, it's used in fisheries biology (estimating fish populations), genetics (inheritance patterns), lottery probabilities, and even in certain aspects of sports analytics.

9. How do you interpret the results of a Hypergeometric probability calculation? The calculated probability (a value between 0 and 1) indicates the likelihood of observing exactly the specified number of successes in a sample drawn without replacement. A higher probability means the event is more likely.

10. What are the differences between the Hypergeometric Distribution and other probability distributions like Binomial? The key difference lies in whether the sampling is done with or without replacement. The Binomial Distribution applies to sampling with replacement (or from an infinite population), where the probability of success remains constant for each trial. The Hypergeometric Distribution applies to sampling without replacement from a finite population, where the probability of success changes with each draw.