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 one success.

Short answer

Answer: The probability of observing exactly one success in this scenario is $\frac{1}{14}$.

Step-by-step solution

Understand the Hypergeometric Probability Formula

The hypergeometric probability formula is used to calculate the probability of observing a specific number of successes (x) in a sample taken from a population without replacement. The formula is given by: $$P(X = k) = \frac{\binom{M}{k} \binom{N - M}{n - k}}{\binom{N}{n}}$$ Here, \(N\) is the population size, \(M\) is the number of successes in the population, \(n\) is the sample size, and \(k\) is the number of successes in the sample.

Identify the Parameters

In this exercise, we are given the following parameters: - Population size \(N = 8\) - Number of successes in the population \(M = 5\) - Sample size \(n = 4\) - Number of successes we want to observe in the sample \(k = 1\)

Apply the Hypergeometric Probability Formula

Using the values of \(N\), \(M\), \(n\), and \(k\), we apply the hypergeometric probability formula: $$P(X = 1) = \frac{\binom{5}{1} \binom{8 - 5}{4 - 1}}{\binom{8}{4}}$$

Calculate the Combinations

Compute the combinations for the formula: - \(\binom{5}{1} = \frac{5!}{1!(5-1)!} = 5\) - \(\binom{3}{3} = \frac{3!}{3!(3-3)!} = 1\) - \(\binom{8}{4} = \frac{8!}{4!(8-4)!} = 70\)

Compute the Probability

Now, substitute the calculated combinations back into the hypergeometric probability formula: $$P(X = 1) = \frac{5 \times 1}{70} = \frac{5}{70}$$ Simplify the fraction to get the probability: $$P(X = 1) = \frac{1}{14}$$ So, the probability of observing exactly one success in a sample of 4 items selected from a population of 8 items, with 5 of these items considered successes, is \(\frac{1}{14}\).

Key concepts

Probability of Observing Successes

Understanding the probability of observing a certain number of successes in a sample is essential in statistical analysis, especially when dealing with discrete variables. This concept revolves around forecasting the likelihood that an event classified as a 'success' will occur a specific number of times under a given set of conditions.

In the context of our exercise, a 'success' refers to drawing an item with a desired characteristic from a larger population (like drawing a winning card from a deck). If we know the total number of successes in the population and the sample size, we can predict how likely it is to observe various counts of successes in our sample.

A critical insight for students is that this probability is not constant; it changes depending on the number of items in the population and the sample, as well as the number of successes one is aiming to find. It's crucial to remember that the larger the sample size or the number of successes in the population, the higher the chance of observing more successes in the sample, up to a certain point.

Combinations in Probability

The concept of combinations is a building block for understanding various probability distributions, including hypergeometric. Combinations tell us how many different ways we can select a group of items from a larger set, without considering the order of selection. If order mattered, we would be discussing permutations instead.

In probability theory, combinations are often represented as 'n choose k,' which quantifies the number of distinct groups of size k that can be picked from n total items. The mathematical representation is written as \( \binom{n}{k} \), and this is calculated using the factorial function (!), where \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).

For students, it is useful to remember that combinations reduce the complexity of calculations. In our exercise, combinations are used to compute the various possible groupings of 'successes' and 'failures' within the sample. It is the factor that bridges raw counts (like '5 successes out of 8') to actual probabilities.

Hypergeometric Distribution

The hypergeometric distribution is a discrete probability distribution that models the number of successes in a series of draws from a finite population without replacement. This last point is key - without replacement means that each draw affects the composition of the population, changing the probabilities with each event.

The classic example used to illustrate the hypergeometric distribution is drawing colored balls from an urn. If you want to know the probability of drawing a certain number of red balls from that urn in a series of draws, you'd use the hypergeometric distribution. The formula heavily relies on combinations to articulate the possible ways of achieving these sets of successful outcomes.

The power of the hypergeometric distribution lies in its ability to model situations with dependent probabilities - where each draw changes the landscape of the population. This differs from the binomial distribution, which models the number of successes in a series of independent events, where the probability remains constant with each draw. It is this difference that shapes the calculations and applications of each distribution.