Hypergeometric Distribution: Videos & Practice Problems

The hypergeometric distribution is a statistical model used when the trials are dependent, meaning the outcome of one trial affects the outcomes of subsequent trials. This contrasts with the binomial distribution, where trials are independent and the probability of success remains constant. In the hypergeometric scenario, we draw items from a finite population without replacement, which alters the composition of the population with each draw.

To illustrate, consider an example where we draw marbles from a bag containing two red and four blue marbles. We want to determine the probability of drawing exactly one red marble in three draws, both with and without replacement. In the case of drawing with replacement, each draw is independent, and the probability of success remains constant at 2 out of 6 (or 1/3). This situation meets the criteria for the binomial distribution.

However, when drawing without replacement, the probability changes with each draw. The hypergeometric distribution is used here, defined by the following parameters:

• n: the number of draws (in this case, 3)
• R: the number of successes in the population (2 red marbles)
• N: the total number of items in the population (6 marbles)

The formula for calculating hypergeometric probabilities is given by:

\[P(X = x) = \frac{{\binom{R}{x} \cdot \binom{N - R}{n - x}}}{{\binom{N}{n}}}\]

In this formula, \(\binom{R}{x}\) represents the number of ways to choose x successes from R successes, while \(\binom{N - R}{n - x}\) represents the number of ways to choose the remaining draws from the failures. The denominator \(\binom{N}{n}\) accounts for the total ways to choose n items from N.

For our example, to find the probability of drawing exactly one red marble, we substitute into the formula:

\[P(X = 1) = \frac{{\binom{2}{1} \cdot \binom{4}{2}}}{{\binom{6}{3}}}\]

Calculating each term, we find:

• \(\binom{2}{1} = 2\)
• \(\binom{4}{2} = 6\)
• \(\binom{6}{3} = 20\)

Thus, the probability becomes:

\[P(X = 1) = \frac{{2 \cdot 6}}{{20}} = \frac{12}{20} = \frac{3}{5}\]

This means the probability of drawing exactly one red marble in three draws without replacement is \(\frac{3}{5}\). Understanding the hypergeometric distribution is crucial for scenarios where the independence of trials cannot be assumed, allowing for accurate probability calculations in dependent situations.

In this scenario, a quality control manager is assessing the reliability of a testing procedure for identifying defective units in a shipment. The shipment consists of 100 units, with 5 known defects. The testing procedure involves randomly selecting 20 units to identify and replace any defective ones. The goal is to ensure that the shipment contains fewer than two defects to avoid complaints from a retail partner.

To determine the probability of successfully identifying and removing enough defective products, we need to find the probability that at least four defects are identified, which would leave at most one defect remaining in the shipment. This can be expressed mathematically as finding the probability that \( x \geq 4 \), where \( x \) represents the number of defects identified.

Using the hypergeometric distribution, we can break this down into two specific probabilities: \( P(x = 4) \) and \( P(x = 5) \). The parameters for our calculations are defined as follows:

• r (the number of successes in the population): 5 (the total number of defective units)
• N (the total number of units in the population): 100
• n (the number of draws): 20 (the number of units tested)

The probability of identifying exactly four defects is calculated using the formula:

\[P(x = 4) = \frac{{\binom{r}{x} \cdot \binom{N - r}{n - x}}}{{\binom{N}{n}}} = \frac{{\binom{5}{4} \cdot \binom{95}{16}}}{{\binom{100}{20}}}\]

Similarly, the probability of identifying all five defects is given by:

\[P(x = 5) = \frac{{\binom{r}{x} \cdot \binom{N - r}{n - x}}}{{\binom{N}{n}}} = \frac{{\binom{5}{5} \cdot \binom{95}{15}}}{{\binom{100}{20}}}\]

By calculating these probabilities and summing them, we find that the total probability of identifying enough defects to avoid a complaint is approximately 0.005, or 0.5%. This indicates a low likelihood of successfully identifying and removing enough defective units.

In part b of the analysis, the quality control manager considers whether to adjust the testing procedure based on the threshold of 10%. Since the calculated probability of 0.5% is significantly below this threshold, it is advisable for the manager to revise the testing procedure to improve the chances of identifying sufficient defects and preventing complaints.