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, 5 are considered "successes." a. Find the probability of observing all successes. b. Find the probability of observing one success. c. Find the probability of observing at most two successes.

Short answer

Answer: The probability of observing at most two successes in the given scenario is \(\frac{1}{2}\).

Step-by-step solution

Calculate Probability of Observing All Successes

To find the probability of observing all successes (\(k=4\)) in a sample of 4 items, we can plug the values into the hypergeometric distribution formula: \(P(X=4) = \frac{\binom{5}{4}\binom{8-5}{4-4}}{\binom{8}{4}}\) Now, compute the binomial coefficients: \(\binom{5}{4} = 5\) \(\binom{8-5}{4-4} = \binom{3}{0} = 1\) \(\bin_pixel{8}{4} = 70\) Finally, plug the binomial coefficients back into the formula and compute the probability: \(P(X=4) = \frac{5 \times 1}{70} = \frac{5}{70} = \frac{1}{14}\) So, the probability of observing all successes in the sample is \(\frac{1}{14}\).

Calculate Probability of Observing One Success

Now, we will calculate the probability of observing one success (\(k=1\)) in the sample: \(P(X=1) = \frac{\binom{5}{1}\binom{8-5}{4-1}}{\binom{8}{4}}\) Computing the binomial coefficients again: \(\binom{5}{1} = 5\) \(\binom{8-5}{4-1} = \binom{3}{3} = 1\) \(\binom{8}{4} = 70\) Plugging in the binomial coefficients, we get: \(P(X=1) = \frac{5 \times 1}{70} = \frac{5}{70} = \frac{1}{14}\) So, the probability of observing one success in the sample is \(\frac{1}{14}\).

Calculate Probability of Observing At Most Two Successes

To find the probability of observing at most two successes (\(k \le 2\)), we can calculate the probabilities for 0, 1, and 2 successes and sum them up: \(P(X \le 2) = P(X=0) + P(X=1) + P(X=2)\) First, find the probability of observing no successes (\(k=0\)): \(P(X=0) = \frac{\binom{5}{0}\binom{8-5}{4-0}}{\binom{8}{4}}\) \(\binom{5}{0}=1\) \(\binom{8-5}{4-0}= \binom{3}{4}=0\) \(\binom{8}{4}=70\) \(P(X=0) = \frac{1 \times 0}{70} = 0\) We already calculated the probability for one success (\(P(X=1) = \frac{1}{14}\)). Now, calculate the probability for two successes (\(k=2\)): \(P(X=2) = \frac{\binom{5}{2}\binom{8-5}{4-2}}{\binom{8}{4}}\) \(\binom{5}{2}=10\) \(\binom{8-5}{4-2}= \binom{3}{2}=3\) \(\binom{8}{4}=70\) \(P(X=2) = \frac{10 \times 3}{70} = \frac{30}{70} = \frac{3}{7}\) Now, we can sum the probabilities for 0, 1, and 2 successes: \(P(X \le 2) = P(X=0) + P(X=1) + P(X=2) = 0 + \frac{1}{14} + \frac{3}{7} = \frac{1}{14} + \frac{6}{14} = \frac{7}{14} = \frac{1}{2}\) So, the probability of observing at most two successes in the sample is \(\frac{1}{2}\).

Key concepts

Probability Calculation

The hypergeometric distribution is a part of probability theory that allows us to calculate the likelihood of a given number of successes in a sample. In this context, a success is one of the favorable outcomes, and a sample is a set of items selected from a larger population.

Here's how you can think about it:

• A population has a certain number of successes and failures.
• We draw a sample and want to find out the probability of observing a specific number of successes.
• The hypergeometric distribution formula considers the number of successes and the sample size.

In this problem, we have a total population of 8 items (5 successes and the rest failures), from which we draw a sample of 4. To find the probability for different numbers of successes, we rely on precise calculations using the formula:\[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \]where:

• \( N \) is the total population size.
• \( K \) is the number of successes in the population.
• \( n \) is the sample size.
• \( k \) is the number of successes in the sample.

The probability, \( P(X = k) \), tells us how likely it is to observe exactly \( k \) successes in our drawn sample.

Binomial Coefficients

Binomial coefficients appear in probability when we are counting combinations of items, and they play a crucial part in the hypergeometric distribution. Binomial coefficients are often written as \( \binom{n}{k} \), which reads as "n choose k." This represents the number of ways to choose \( k \) successes from \( n \) items without regard to order.

It's calculated using the formula:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

where \( n! \) is the factorial of \( n \), meaning the product of all positive integers up to \( n \). Let's break this down a bit:

• The factorial symbol \(!\) means you multiply the number by every number less than it, down to 1.
• For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
• The binomial coefficient \( \binom{5}{2} \) means you want the number of ways to choose 2 successes from a group of 5.

For the problem we solved, binomial coefficients like \( \binom{5}{4} \) and \( \binom{8}{4} \) help determine the number of ways to select items in our hypergeometric distribution problems. Understanding this gives clarity on how each success configuration is part of our probability calculations.

Combinatorial Analysis

Combinatorial analysis is the mathematical study of counting and arrangement. In our specific case, it helps us solve problems of selecting a group from a larger set. This analysis is central to solving problems involving hypergeometric distribution.

Think of combinatorial analysis as a toolkit to answer questions like "How many ways can we pick 4 items from 8?" When tasked with finding the probability of observing a certain configuration of successes in a sample, combinatorial analysis becomes indispensable.

With combinatorial analysis, you can:

• Identify the number of outcomes that fit a specific set of conditions (like observing all successes).
• Determine all possible combinations of outcomes (some combination leading to more, fewer, or exact successes).
• Apply these to calculate probabilities in complex situations by counting different combinations effectively.

In our case, we use combinatorial analysis to enumerate and evaluate all possible ways to select our sample of successes for precise probability computations. This enriches our understanding of how different scenarios stack up against each other in terms of likelihood.