Question

Let \(x\) be the number of successes observed in a sample of \(n=4\) items selected from a population of \(N=8 .\) Suppose that of the \(N=8\) items, \(M=5\) are considered "successes." Find the probabilities in Exercises \(8-10 .\) The probability of observing all successes.

Short answer

Answer: The probability of observing all 4 successes in the given sample is $\frac{1}{14}$.

Step-by-step solution

Write down the known values

In this problem, we have the following values given: - Population size, \(N=8\) - Number of successes in the population, \(M=5\) - Sample size, \(n=4\) - Number of successes in the sample, \(x=4\)

Use the hypergeometric distribution formula to calculate the probability

Using the formula for the hypergeometric distribution: \(P(X=x) = \frac{\binom{M}{x} \binom{N-M}{n-x}}{\binom{N}{n}}\) Substitute the given values into the formula: \(P(X=4) = \frac{\binom{5}{4} \binom{8-5}{4-4}}{\binom{8}{4}}\)

Calculate the combinations

Calculate the combinations in the numerator and the denominator: \(\binom{5}{4} = \frac{5!}{4!(5-4)!} = 5\) \(\binom{8-5}{4-4} = \binom{3}{0} = \frac{3!}{0!(3-0)!} = 1\) \(\binom{8}{4} = \frac{8!}{4!(8-4)!} = 70\)

Calculate the probability

Now substitute the combinations back into the formula: \(P(X=4) = \frac{5 \cdot 1}{70} = \frac{5}{70}\) Simplify the fraction: \(P(X=4) = \frac{1}{14}\) Therefore, the probability of observing all successes in the sample of \(n=4\) items selected from the population of \(N=8\) items with \(M=5\) successes is \(\boxed{\frac{1}{14}}\).

Key concepts

Probability of Successes

Probability refers to the likelihood of an event occurring. In the context of the hypergeometric distribution, the 'probability of successes' is the chance of observing a certain number of successes in a sample drawn without replacement from a finite population.

For instance, if you're picking 4 cards (\(n=4\)) from a standard deck of 52 cards, where there are 20 red cards (\(M=20\)), what's the probability that all four cards you pick will be red? This scenario illustrates the 'probability of successes' in action. By harnessing the hypergeometric distribution, we can calculate exact probabilities rather than relying on approximations that would be necessary if we were sampling with replacement.

Understanding this concept helps students tackle problems where they are required to determine how likely it is to get a certain number of successful outcomes from a sample drawn from a population with a known number of successes.

Combinatorial Calculations

Combinatorial calculations are mathematical techniques used to count or enumerate various ways of arranging or selecting objects without actually listing them. Central to these calculations is the concept of combinations, denoted as \(\binom{n}{k}\) and spoken as 'n choose k.' It represents the number of ways to choose a sample of size \(k\) from a larger set of size \(n\), without regard to the order.

The formula for a combination is \(\binom{n}{k} = \frac{n!}{k! (n-k)!}\), where \(n!\) denotes the factorial of \(n\). Factorials are a sequence where \(n! = n \times (n-1) \times (n-2) \times ... \times 1\), with the specific case of \(0! = 1\).

In the exercise provided, combinatorial calculations allow us to determine the number of ways to select 4 successes from 5 (\(\binom{5}{4}\)), and they are a critical step in arriving at the probability of observing all successes with the hypergeometric distribution. Without these calculations, we cannot accurately calculate probabilities for complex scenarios.

Statistical Distributions

Statistical distributions are fundamental concepts in statistics that describe how probabilities are spread out across different outcomes. They give us tools to model and understand real-world phenomena, ranging from natural events to daily occurrences.

The hypergeometric distribution is a discrete probability distribution that describes the odds for a specific number of successes in a series of draws from a finite population without replacement. It is distinct from other distributions like the binomial distribution (which permits replacement) and the normal distribution (which is continuous and used for different scenarios).

Our exercise makes use of the hypergeometric distribution to solve for the probability of drawing all successes in a sample. Recognizing the type of distribution to use is essential as it influences how we approach problem-solving in statistics. Grasping the intricacies of each distribution empowers students to analyze data correctly and make predictions with confidence.