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Suppose \(N=10\) and \(r=3 .\) Compute the hypergeometric probabilities for the following values of \(n\) and \(x\) a. \(\quad n=4, x=1\) b. \(n=2, x=2\) c. \(\quad n=2, x=0\) d. \(n=4, x=2\) e. \(n=4, x=4\)

Short Answer

Expert verified
a) 0.5, b) 0.0667, c) 0.4667, d) 0.3, e) 0

Step by step solution

01

Understand the Hypergeometric Distribution

The hypergeometric distribution describes the probability of drawing a specific number of successes from a finite population without replacement. For any draw, the probability is defined by the formula: \\[ P(X = x) = \frac{{\binom{r}{x} \times \binom{N-r}{n-x}}}{\binom{N}{n}} \] \where \(N\) is the total population size, \(r\) is the number of successes in the population, \(n\) is the number of draws, and \(x\) is the number of observed successes in those draws.
02

Step 2a: Calculate Hypergeometric Probability for n=4, x=1

Given \(N=10\), \(r=3\), \(n=4\), and \(x=1\), apply the hypergeometric formula: \\[ P(X = 1) = \frac{{\binom{3}{1} \times \binom{7}{3}}}{\binom{10}{4}} \] \Calculate the binomial coefficients: \\(\binom{3}{1} = 3\), \(\binom{7}{3} = 35\), \(\binom{10}{4} = 210\). Substitute these into the equation: \\[ P(X = 1) = \frac{3 \times 35}{210} = \frac{105}{210} = 0.5 \]
03

Step 2b: Calculate Hypergeometric Probability for n=2, x=2

Given \(N=10\), \(r=3\), \(n=2\), and \(x=2\), apply the formula: \\[ P(X = 2) = \frac{{\binom{3}{2} \times \binom{7}{0}}}{\binom{10}{2}} \] \Calculate the binomial coefficients: \\(\binom{3}{2} = 3\), \(\binom{7}{0} = 1\), \(\binom{10}{2} = 45\). Substitute these into the equation: \\[ P(X = 2) = \frac{3 \times 1}{45} = \frac{3}{45} = 0.0667 \]
04

Step 2c: Calculate Hypergeometric Probability for n=2, x=0

Given \(N=10\), \(r=3\), \(n=2\), and \(x=0\), apply the formula: \\[ P(X = 0) = \frac{{\binom{3}{0} \times \binom{7}{2}}}{\binom{10}{2}} \] \Calculate the binomial coefficients: \\(\binom{3}{0} = 1\), \(\binom{7}{2} = 21\), \(\binom{10}{2} = 45\). Substitute these into the equation: \\[ P(X = 0) = \frac{1 \times 21}{45} = \frac{21}{45} = 0.4667 \]
05

Step 2d: Calculate Hypergeometric Probability for n=4, x=2

Given \(N=10\), \(r=3\), \(n=4\), and \(x=2\), apply the formula: \\[ P(X = 2) = \frac{{\binom{3}{2} \times \binom{7}{2}}}{\binom{10}{4}} \] \Calculate the binomial coefficients: \\(\binom{3}{2} = 3\), \(\binom{7}{2} = 21\), \(\binom{10}{4} = 210\). Substitute these into the equation: \\[ P(X = 2) = \frac{3 \times 21}{210} = \frac{63}{210} = 0.3 \]
06

Step 2e: Calculate Hypergeometric Probability for n=4, x=4

Given \(N=10\), \(r=3\), \(n=4\), and \(x=4\), since \(x\) cannot be greater than \(r\), the probability is 0.

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Key Concepts

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

Hypergeometric Probability
The hypergeometric probability refers to the probability of a given number of successes in a set of draws, without replacement, from a finite population. This is different from the binomial distribution, where each draw is independent and with replacement. In the hypergeometric distribution, the result of each draw affects the remaining population. This probability is calculated using the hypergeometric formula:
\[P(X = x) = \frac{{\binom{r}{x} \times \binom{N-r}{n-x}}}{\binom{N}{n}}\]
Let's break this formula down:- **\(N\)** is the total population size.- **\(r\)** is the total number of successes in the population.- **\(n\)** is the number of draws.- **\(x\)** is the number of observed successes.
The top part of the fraction uses binomial coefficients to account for the ways of selecting successes and failures, while the bottom part, \(\binom{N}{n}\), denotes the total possible combinations of drawing \(n\) items from \(N\). The hypergeometric probability is particularly helpful in scenarios like quality control, where sampling is done from a finite group.
Binomial Coefficients
Binomial coefficients are central to the calculation of hypergeometric probabilities. They are used to determine the number of ways to choose a given number of elements from a larger set. The notation \(\binom{n}{k}\) reads as "n choose k" and is defined as:
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]
Where "!" denotes a factorial—meaning to multiply the sequence of descending natural numbers. Factorials are a product of intuition and calculation, helping compute combinations effectively. For example, choosing 2 successes out of 3 in the population is calculated as \(\binom{3}{2} = 3\). Binomial coefficients simplify the process of finding successful draws in complicated probability distributions.

These coefficients are also symmetric; \(\binom{n}{k} = \binom{n}{n-k}\), which means the number of ways to choose \(k\) items is the same as the number of ways to leave \(k\) items out.
Finite Population Sampling
Finite population sampling is a statistical method where samples are drawn from a limited or fixed number of elements. Unlike infinite population sampling, finite sampling explicitly acknowledges the limitations on the number of units available. This is vital in the context of hypergeometric probability because it directly influences the outcomes of interest.

In hypergeometric distribution, each draw affects the subsequent draws since elements are not replaced. A finite sample reflects real-world limitations more accurately, such as when testing to find defects in a limited batch of items. It's crucial to understand that this lack of replacement leads to dependent events.
  • **Without Replacement:** Once an element is drawn, it is not returned to the population.
  • **Finite Group Context:** Decisions based on finite sampling are often used in quality assurance and ecological studies because they offer a realistic model for resource-constrained conditions.
Ultimately, understanding finite population sampling helps in making informed statistical inferences in real-life scenarios.

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Most popular questions from this chapter

The National Basketball Association (NBA) records a variety of statistics for each team. Two of these statistics are the percentage of field goals made by the team and the percentage of three-point shots made by the team. For a portion of the 2004 season, the shooting records of the 29 teams in the NBA showed that the probability of scoring two points by making a field goal was \(.44,\) and the probability of scoring three points by making a threepoint shot was .34 (NBA website, January 3,2004 ). a. What is the expected value of a two-point shot for these teams? b. What is the expected value of a three-point shot for these teams? c. If the probability of making a two-point shot is greater than the probability of making a three-point shot, why do coaches allow some players to shoot the three-point shot if they have the opportunity? Use expected value to explain your answer.

Blackjack, or twenty-one as it is frequently called, is a popular gambling game played in Las Vegas casinos. A player is dealt two cards. Face cards (jacks, queens, and kings) and tens have a point value of \(10 .\) Aces have a point value of 1 or \(11 .\) A 52 -card deck contains 16 cards with a point value of 10 (jacks, queens, kings, and tens) and four aces. a. What is the probability that both cards dealt are aces or 10 -point cards? b. What is the probability that both of the cards are aces? c. What is the probability that both of the cards have a point value of \(10 ?\) d. \(A\) blackjack is a 10 -point card and an ace for a value of \(21 .\) Use your answers to parts (a), (b), and (c) to determine the probability that a player is dealt blackjack. (Hint: Part (d) is not a hypergeometric problem. Develop your own logical relationship as to how the hypergeometric probabilities from parts (a), (b), and (c) can be combined to answer this question.)

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