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Let \(x\) be the number of successes observed in a sample of \(n=5\) items selected from \(N=10 .\) Suppose that, of the \(N=10\) items, 6 are considered "successes." a. Find the probability of observing no successes. b. Find the probability of observing at least two successes. c. Find the probability of observing exactly two successes.

Short Answer

Expert verified
Answer: The probability of observing no successes is \(P(X=0) = 0\). 2. What is the probability of observing at least two successes in a sample of 5 items from a population of 10 items, where 6 are considered successes? Answer: The probability of observing at least two successes is \(P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5)\). Specific values of each probability should be calculated and summed up to get the final probability. 3. What is the probability of observing exactly two successes in a sample of 5 items from a population of 10 items, where 6 are considered successes? Answer: The probability of observing exactly two successes is \(P(X=2) = \frac{\binom{6}{2} \binom{10 - 6}{5-2}}{\binom{10}{5}}\).

Step by step solution

01

a. Probability of observing no successes

We need to find the probability of having \(0\) successes in our sample, i.e., \(P(X=0)\). Using the hypergeometric probability formula: \(P(X=0) = \frac{\binom{6}{0} \binom{10-6}{5-0}}{\binom{10}{5}} = \frac{\binom{6}{0} \binom{4}{5}}{\binom{10}{5}}\) Notice that \(\binom{4}{5}=0\) because, in this case, we are trying to select \(5\) items from a group of \(4\) which is not possible. Therefore, \(P(X=0) = \frac{\binom{6}{0} \times 0}{\binom{10}{5}} = 0\)
02

b. Probability of observing at least two successes

To find the probability of observing at least two successes, we can sum up the probabilities of observing \(2\), \(3\), \(4\), and \(5\) successes. That is, \(P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5)\). We can calculate each probability separately and then add them up: \(P(X=2) = \frac{\binom{6}{2} \binom{10 - 6}{5-2}}{\binom{10}{5}}\) \(P(X=3) = \frac{\binom{6}{3} \binom{10 - 6}{5-3}}{\binom{10}{5}}\) \(P(X=4) = \frac{\binom{6}{4} \binom{10 - 6}{5-4}}{\binom{10}{5}}\) \(P(X=5) = \frac{\binom{6}{5} \binom{10 - 6}{5-5}}{\binom{10}{5}}\) Now, we can add up the probabilities: \(P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5)\)
03

c. Probability of observing exactly two successes

We already found the probability of observing exactly two successes, which we calculated while solving part b. So, the probability of observing exactly two successes is given as: \(P(X=2) = \frac{\binom{6}{2} \binom{10 - 6}{5-2}}{\binom{10}{5}}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability of Observing Successes
Understanding the probability of observing a certain number of successes in a specific scenario is crucial in statistics. It allows us to predict how likely it is to obtain a result we deem as 'successful' given certain conditions. In the context of our exercise, a success refers to selecting an item that meets our criteria of interest from a finite population.

In our example, we're dealing with a hypergeometric distribution, which applies when we're drawing without replacement from a finite population. There are 6 'successes' among 10 items, and we want to know the likelihood of drawing a certain number of these successes in a sample of 5 items. The notion of 'successes' could relate to anything from defective products in a batch to having a certain gene in a group of organisms. The application of this concept is wide-reaching in fields as diverse as quality control, biology, and even game theory.
Hypergeometric Probability Formula
The hypergeometric probability formula is essential for solving problems where you are randomly selecting items without replacement from a finite population. The formula is given by:\[P(X=k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{{\binom{N}{n}}}\]Where:
  • \(N\) is the total number of items in the population.
  • \(K\) is the number of successes in the population.
  • \(n\) is the number of items drawn from the population.
  • \(k\) is the number of observed successes.
  • \(\binom{n}{k}\) is a binomial coefficient, calculating the number of ways to choose \(k\) successes from \(n\) items.

Applying this formula requires understanding of combinatorics and recognizing the constraints of drawing without replacement. The power of this formula lies in its ability to calculate exact probabilities for complex draw scenarios which are quite common in various real-world applications.
Combinatorics in Probability
Combinatorics plays a major role in probability, particularly in calculating the numerous ways certain events can occur. One of the fundamental tools of combinatorics is the binomial coefficient, often read as 'n choose k', represented by \(\binom{n}{k}\). This coefficient calculates the number of combinations, or ways to choose \(k\) items out of \(n\), disregarding the order of selection.

In probability calculations, combinatorics helps quantify the various possibilities. For instance, it tells us how many different samples of a certain size we can draw from a population. The hypergeometric distribution relies on these combinatorial calculations to find probabilities of different outcomes when sampling without replacement as seen in our exercise.
Probability Calculations
To solve probability problems, it's often necessary to perform calculations involving binomial coefficients and probability formulas. These calculations can involve summing the probabilities of multiple mutually exclusive events or using formulas like the hypergeometric distribution to find the probability of a single event. It's crucial to approach these calculations methodically.

For example, the exercise illustrates a calculation for the probability of drawing at least two successes from the population. This requires adding the probability of obtaining exactly two, three, four, and five successes. Each of these probabilities is calculated using the hypergeometric formula, which accounts for different numbers of successes and failures in the sample. Accurate probability calculations are the foundation of making informed predictions and decisions based on a set of given data.

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