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Let \(N=16, r=10\), and \(n=5\). Using the hypergeometric probability distribution formula, find a. \(P(x=5)\) b. \(P(x=0)\) c. \(P(x \leq 1)\)

Short Answer

Expert verified
a. The probability \(P(x = 5)\) is approximately 0.058. b. The probability \(P(x = 0)\) is approximately 0.0014. c. The probability \(P(x \leq 1)\) is approximately 0.0358.

Step by step solution

01

Calculation for \(P(x=5)\)

Substitute \(N=16\), \(r=10\), \(n=5\), and \(x=5\) into the formula to get\[P(5)= \dfrac{\binom{10}{5}\binom{16 - 10}{5 - 5}}{\binom{16}{5}}\]Next, working with the binomial coefficients results in\[P(5)= \dfrac{\binom{10}{5}\binom{6}{0}}{\binom{16}{5}}\]which simplifies to\[P(5)= \dfrac{252 \times 1}{4368} = 0.058\]
02

Calculation for \(P(x=0)\)

Substitute \(N=16\), \(r=10\), \(n=5\), and \(x=0\) into the formula to get\[P(0)= \dfrac{\binom{10}{0}\binom{16 - 10}{5 - 0}}{\binom{16}{5}}\]Next, working with the binomial coefficients results in\[P(0)= \dfrac{\binom{10}{0}\binom{6}{5}}{\binom{16}{5}}\]which simplifies to\[P(0)= \dfrac{1 \times 6}{4368} = 0.0014\]
03

Calculation for \(P(x \leq 1)\)

For this part, you need to find the probabilities for \(x=0\) and \(x=1\) and then sum them up. We already know the value for \(x=0\), so you need to calculate for \(x=1\).Substitute \(N=16\), \(r=10\), \(n=5\), and \(x=1\) into the formula to get\[P(1)= \dfrac{\binom{10}{1}\binom{16 - 10}{5 - 1}}{\binom{16}{5}}\]Next, working with the binomial coefficients results in\[P(1)= \dfrac{\binom{10}{1}\binom{6}{4}}{\binom{16}{5}}\]which simplifies to\[P(1)= \dfrac{10 \times 15}{4368} = 0.0344\]Therefore, \(P(x \leq 1)=P(0)+P(1)=0.0014+0.0344=0.0358\)

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Key Concepts

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

Probability Calculation
To understand how the probability calculation works in the context of hypergeometric distribution, let’s break it down step by step. Imagine a scenario where you have a box of colored balls, and you know how many balls of each color are inside. Suppose you want to know the likelihood of drawing a specific number of a certain color when you draw a few balls at random. That's essentially what you're calculating with hypergeometric distribution.

In this context, the general formula used is:
\[P(x) = \frac{\binom{r}{x} \binom{N-r}{n-x}}{\binom{N}{n}}\]Here,
  • \(N\) is the total number of items,
  • \(r\) is the number of success states in the population,
  • \(n\) is the number of draws,
  • \(x\) is the number of observed successes.
This formula calculates the probability of drawing exactly \(x\) successes from \(n\) draws, without replacement. Each probability calculation step involves substituting the specific values into this formula, performing binomial coefficient computations, and simplifying the results.
Binomial Coefficients
Binomial coefficients, sometimes called combinations, play a crucial role in probability calculations involving hypergeometric distribution. The notation \(\binom{a}{b}\) represents the number of ways to choose \(b\) items from \(a\) items without regard to order. It is calculated using the formula:
\[\binom{a}{b} = \frac{a!}{b!(a-b)!}\]where \(a!\) denotes factorial of \(a\).

In the context of our exercise, binomial coefficients are used to determine the number of ways to select certain combinations of successes and failures when calculating probabilities. For example, in calculating \(P(x=5)\), the binomial coefficients are used to find the ways to pick 5 successful outcomes from 10 possible successes and none from the remaining failures. These coefficients simplify to determine feasible selections without replacement, showcasing scenarios where each draw directly impacts subsequent outcomes, differing from binomial distribution where such dependencies do not exist.
Statistical Distributions
Statistical distributions characterize how data points, like probabilities, are spread across a possible range of outcomes. The hypergeometric distribution is one type of statistical distribution that applies to scenarios where sampling is done without replacement.
  • Unlike binomial distribution which assumes replacement, each draw affects the next in hypergeometric distribution.
  • It’s useful when considering random events in small populations or scenarios where every element has a significant impact.
This distribution is defined by its three central parameters: total population size \(N\), number of successes in the population \(r\), and the number of draws \(n\). The hypergeometric distribution is particularly relevant in quality control, lotteries, or any activity where the probability of success changes with each draw due to the lack of replacement.

Understanding these statistical distributions provides insights into the probabilistic landscape and assists in making informed decisions based on statistical variability and outcomes. Whether you're analyzing quality in manufacturing or predicting outcomes in sampling surveys, appreciating how each draw interacts with the prior helps in accurate probability evaluation.

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