Hypergeometric Distribution: Videos & Practice Problems

The hypergeometric distribution is used when trials are dependent, particularly in scenarios where items are drawn without replacement. Unlike the binomial distribution, where trials are independent, the hypergeometric model adjusts probabilities based on previous draws. For example, when drawing marbles from a bag, the probability of drawing a specific number of successes changes as items are removed. The formula for hypergeometric probabilities involves combinations to determine the likelihood of achieving a desired number of successes in a fixed number of draws from a finite population.

Introduction to the Hypergeometric Distribution

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 have a constant probability of success. 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. Here, we can apply the binomial distribution formula:

For a binomial distribution, the probability of getting exactly \( x \) successes in \( n \) trials is given by:

\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \]

Where \( p \) is the probability of success on each trial. In our example, the probability of drawing one red marble with replacement simplifies to \( \frac{4}{9} \).

However, when drawing without replacement, we must use the hypergeometric distribution. The relevant parameters are:

• \( N \): Total number of items (6 marbles)
• \( n \): Number of draws (3)
• \( R \): Number of successes in the population (2 red marbles)
• \( r \): Number of successes we want (1 red marble)

The hypergeometric probability formula is expressed as:

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

In our case, we calculate:

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

Calculating each term, we find:

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

Thus, the probability becomes:

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

This result indicates that the probability of drawing exactly one red marble in three draws without replacement is \( \frac{3}{5} \). Understanding the differences between the binomial and hypergeometric distributions is crucial for accurately modeling scenarios where the independence of trials cannot be assumed.

A school is holding a fair raffle and a teacher is interested in predicting how many winners will be from her class. Determine which probability distribution she should use given the following information.
(A) There are 386 tickets, one for each student. Tickets are placed back in the pool after being chosen and 5 tickets are drawn.

• A. Binomial
• B. Hypergeometric

Answer: (A)

A school is holding a fair raffle and a teacher is interested in predicting how many winners will be from her class. Determine which probability distribution she should use given the following information.
(B) There are 386 tickets, one for each student. Tickets are removed from the pool after being chosen and 5 tickets are drawn.

• A. Binomial
• B. Hypergeometric

Answer: (B)

Find the probability of drawing a hand of 5 cards from a standard deck that contains exactly 2 hearts.

• A. $0.15$
• B. $0.73$
• C. $0.038$
• D. $0.27$

Answer: (D)

Introduction to the Hypergeometric Distribution

In this scenario, a quality control manager is assessing the reliability of a testing procedure to identify 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 the random variable \(X\) (the number of defects identified) is greater than or equal to 4, or \(P(X \geq 4)\). This can be broken down into two specific probabilities: \(P(X = 4)\) and \(P(X = 5)\).

To apply the hypergeometric distribution, we define the following parameters:

• 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 4 defects is calculated using the formula:

\[P(X = 4) = \frac{{\binom{r}{4} \cdot \binom{N - r}{n - 4}}}{{\binom{N}{n}}}\]

Substituting the values, we have:

\[P(X = 4) = \frac{{\binom{5}{4} \cdot \binom{95}{16}}}{{\binom{100}{20}}}\]

For the probability of identifying all 5 defects, the formula is:

\[P(X = 5) = \frac{{\binom{r}{5} \cdot \binom{N - r}{n - 5}}}{{\binom{N}{n}}}\]

Substituting the values, we have:

\[P(X = 5) = \frac{{\binom{5}{5} \cdot \binom{95}{15}}}{{\binom{100}{20}}}\]

Adding these probabilities together gives us:

\[P(X \geq 4) = P(X = 4) + P(X = 5)\]

Upon calculating these probabilities, the result is approximately 0.005, indicating a 0.5% chance that the testing procedure will identify enough defects to avoid a complaint.

In part b, the quality control manager considers adjusting the testing procedure if the probability of successfully removing enough defective units falls below 10%. Since 0.5% is significantly less than 10%, it is advisable for the manager to revise the testing procedure to improve the chances of identifying and removing defects effectively.

Here’s what students ask on this topic:

What is the difference between the hypergeometric distribution and the binomial distribution?

The hypergeometric distribution differs from the binomial distribution primarily in the dependency of trials. In the binomial distribution, trials are independent, meaning the outcome of one trial does not affect the probability of success in subsequent trials. The probability of success remains constant throughout. In contrast, the hypergeometric distribution deals with dependent trials, where items are drawn without replacement. This causes the probability of success to change as items are removed from the population. For example, if you draw marbles from a bag without replacement, the composition of the bag changes after each draw, altering the probability of success for future draws. This dependency is the key distinction between the two distributions.

How do you calculate probabilities using the hypergeometric distribution formula?

The hypergeometric distribution formula calculates the probability of obtaining a specific number of successes in a fixed number of draws without replacement. The formula is:

$\frac{\left(R_x^{}\right)\left(N_n^{}\right)}{N_n^{}}$

Here, $R$ is the total number of successes in the population, $N$ is the total population size, $n$ is the number of draws, and $x$ is the number of successes desired. The numerator calculates the number of ways to achieve the desired successes and failures, while the denominator calculates the total possible draws. Simplify the combinations using factorials or a calculator for practical computation.

What are the conditions for using the hypergeometric distribution?

The hypergeometric distribution is applicable under three specific conditions: (1) The population is finite and consists of two distinct groups, such as successes and failures. (2) Items are drawn without replacement, meaning the population size decreases after each draw. (3) The number of draws is fixed. These conditions ensure that the trials are dependent, and the probability of success changes as items are removed from the population. For example, drawing cards from a deck without replacement meets these conditions, as the composition of the deck changes after each draw.

Can you provide an example of a hypergeometric probability calculation?

Consider a bag with 6 marbles: 2 red (successes) and 4 blue (failures). You draw 3 marbles without replacement and want the probability of getting exactly 1 red marble. Using the hypergeometric formula:

$\frac{\left(2_1^{}\right)\left(4_2^{}\right)}{6_3^{}}$

Here, $R=\; 2$, $N=\; 6$, $n=\; 3$, and $x=\; 1$. Simplify the combinations: $2_1^{}=\; 2$, $4_2^{}=\; 6$, and $6_3^{}=\; 20$. The probability is:

$\frac{12}{20}$

or 3/5.

Why is the hypergeometric distribution important in statistics?

The hypergeometric distribution is crucial for modeling scenarios where trials are dependent, such as sampling without replacement. It accurately reflects real-world situations where the probability of success changes as items are removed from the population. This makes it valuable in fields like quality control, biology, and social sciences, where finite populations are studied. For example, in quality control, it can determine the likelihood of defective items in a sample without testing the entire batch. Understanding this distribution helps statisticians analyze dependent events and make informed decisions based on changing probabilities.