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Suppose that \(X\) has a hypergeometric distribution with \(N=10, n=3,\) and \(K=4\). Sketch the probability mass function of \(X\). Determine the cumulative distribution function for \(X\).

Short Answer

Expert verified
The PMF is graphed with values 0, 1, 2, 3 having probabilities 1/6, 1/2, 3/10, and 1/30, respectively. The CDF increases as cumulative probabilities 1/6, 2/3, 19/30, 1.

Step by step solution

01

Identify the Hypergeometric Distribution Formula

The probability mass function (PMF) for a hypergeometric distribution is given by \[ P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{\binom{N}{n}} \]where \( N \) is the population size, \( K \) is the number of successes in the population, \( n \) is the number of draws, and \( k \) is the number of observed successes. For this problem, \( N = 10 \), \( n = 3 \), and \( K = 4 \).
02

Calculate Probability for Each Possible Value of X

The possible values of \( X \) range from 0 to 3 since \( n = 3 \). Calculate the probabilities for each \( X = k \).- For \( k = 0 \): \[ P(X = 0) = \frac{{\binom{4}{0} \binom{6}{3}}}{\binom{10}{3}} = \frac{1 \times 20}{120} = \frac{1}{6} \]- For \( k = 1 \): \[ P(X = 1) = \frac{{\binom{4}{1} \binom{6}{2}}}{\binom{10}{3}} = \frac{4 \times 15}{120} = \frac{1}{2} \]- For \( k = 2 \): \[ P(X = 2) = \frac{{\binom{4}{2} \binom{6}{1}}}{\binom{10}{3}} = \frac{6 \times 6}{120} = \frac{3}{10} \]- For \( k = 3 \): \[ P(X = 3) = \frac{{\binom{4}{3} \binom{6}{0}}}{\binom{10}{3}} = \frac{4 \times 1}{120} = \frac{1}{30} \]
03

Sketch the Probability Mass Function (PMF)

Using the probabilities calculated, the PMF graph can be sketched. It plots the values of \( X \) on the horizontal axis (0, 1, 2, 3) and their corresponding probabilities on the vertical axis (\( \frac{1}{6}, \frac{1}{2}, \frac{3}{10}, \frac{1}{30} \)).
04

Calculate the Cumulative Distribution Function (CDF)

The cumulative distribution function \( F(x) \) provides the probability that \( X \leq x \). It is calculated by summing the probabilities:- For \( x = 0 \): \[ F(0) = P(X \leq 0) = \frac{1}{6} \]- For \( x = 1 \): \[ F(1) = P(X \leq 1) = P(X = 0) + P(X = 1) = \frac{1}{6} + \frac{1}{2} = \frac{2}{3} \]- For \( x = 2 \): \[ F(2) = P(X \leq 2) = F(1) + P(X = 2) = \frac{2}{3} + \frac{3}{10} = \frac{19}{30} \]- For \( x = 3 \): \[ F(3) = P(X \leq 3) = F(2) + P(X = 3) = \frac{19}{30} + \frac{1}{30} = 1 \]

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (PMF) is an essential concept in understanding discrete probability distributions, such as the hypergeometric distribution. It gives us the probability that a discrete random variable, such as \( X \), takes on a particular value \( k \).
  • In the case of a hypergeometric distribution, the PMF is calculated using the formula: \[ P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}}{\binom{N}{n}} \]
  • \( N \) is the population size.
  • \( K \) is the number of successes in that population.
  • \( n \) is the number of draws or samples taken.
  • \( k \) is the number of observed successes within those samples.
The PMF helps us construct a probability distribution by listing all possible values of \( X \) along with their probabilities. In the given example, the various values for \( k \) can be 0, 1, 2, and 3, because the number of chosen elements \( n \) is 3. This structured method ensures that every outcome is covered and helps us visualize how probable each scenario is.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) complements the PMF by showing the cumulative probability of a variable being less than or equal to a certain value.
  • The CDF, denoted as \( F(x) \), is calculated by summing the probabilities from the PMF for all values less than or equal to \( x \).
  • Mathematically, it is expressed as: \[ F(x) = P(X \leq x) \]
For example, to find \( F(2) \), you would add the probabilities for \( X = 0, 1, \) and \( 2 \). This gives us a comprehensive view of the likelihood that an outcome is within or below a certain threshold.
In the hypergeometric distribution example, \( F(0) = \frac{1}{6} \), representing the probability that zero successes are observed. As we calculate higher values, we see each step reflect the total probability accumulated, thus illustrating shifts in probability distribution as we move across possible outcomes.
Combinatorics
Combinatorics plays a vital role in probability and statistics, especially when calculating probabilities for hypergeometric distributions. This field of mathematics focuses on counting and ordering possibilities, something often essential in probability scenarios.
  • The key combinatorial function used is the binomial coefficient, represented by \( \binom{n}{k} \), which calculates the number of ways \( k \) items can be chosen from \( n \) items without regard to order.
  • In the PMF for a hypergeometric distribution, combinatorics helps determine the number of favorable outcomes over the total number of outcomes possible.
The formula for computing the binomial coefficient is:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]Where \( ! \) denotes factorial, meaning the product of all positive integers up to that number.
In the hypergeometric problem, combinatorics is applied to both the numerator and denominator of the PMF formula. By calculating combinations for both the successes and the remainder of the population, we refine our understanding of how each scenario fits within the total framework of possible outcomes.

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