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Let \(x\) be the number of successes observed in a sample of \(n=5\) items selected from a population of \(N=10 .\) Suppose that of the \(N=10\) items, \(M=6\) are considered "successes." Find the probabilities in Exercises \(11-13 .\) The probability of observing at least two successes.

Short Answer

Expert verified
Answer: The probability of observing at least two successes is 246/252.

Step by step solution

01

Understand the hypergeometric probability formula

The hypergeometric probability formula is as follows: P(X=k) = \frac{\binom{M}{k}\binom{N-M}{n-k}}{\binom{N}{n}} where - N = the total number of items in the population - M = the number of successes in the population - n = the number of items in the sample - k = the number of successes in the sample In this case, we have N = 10, M = 6, and n = 5.
02

Calculate the probability for each number of successes

We need to find the probability of observing at least two successes, which means we need to find the probabilities for k = 2, 3, 4, and 5 successes. P(X=2) = \frac{\binom{6}{2}\binom{4}{3}}{\binom{10}{5}} P(X=3) = \frac{\binom{6}{3}\binom{4}{2}}{\binom{10}{5}} P(X=4) = \frac{\binom{6}{4}\binom{4}{1}}{\binom{10}{5}} P(X=5) = \frac{\binom{6}{5}\binom{4}{0}}{\binom{10}{5}}
03

Evaluate the probabilities

Now, calculate the binomial coefficients and probabilities for each case: P(X=2) = \frac{\binom{6}{2}\binom{4}{3}}{\binom{10}{5}} = \frac{15\cdot 4}{252} = \frac{60}{252} P(X=3) = \frac{\binom{6}{3}\binom{4}{2}}{\binom{10}{5}} = \frac{20\cdot 6}{252} =\frac{120}{252} P(X=4) = \frac{\binom{6}{4}\binom{4}{1}}{\binom{10}{5}} = \frac{15\cdot 4}{252} = \frac{60}{252} P(X=5) = \frac{\binom{6}{5}\binom{4}{0}}{\binom{10}{5}} = \frac{6\cdot 1}{252} = \frac{6}{252}
04

Find the probability of observing at least two successes

Add up the probabilities for k = 2, 3, 4, and 5 to get the probability of observing at least two successes: P(X >= 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5) = \frac{60}{252} + \frac{120}{252} + \frac{60}{252} + \frac{6}{252} = \frac{246}{252} Therefore, the probability of observing at least two successes is \frac{246}{252}.

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Key Concepts

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

Binomial Coefficients
Binomial coefficients are integral components of combinatorics and probability theory. They are often written in the form \( \binom{n}{k} \) and pronounced 'n choose k'. This notation represents the number of distinct ways to choose \( k \) items from a total of \( n \) without concern for the arrangement of those items.

For instance, if we were forming committees from a group of people, \( \binom{n}{k} \) would tell us how many possible committees of \( k \) members could exist. Mathematically, these coefficients are defined using factorials: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), where \( n! \) denotes the factorial of \( n \) – the product of all positive integers up to \( n \).

Understanding binomial coefficients is pivotal when working with distributions that involve combinations, like the hypergeometric distribution in our exercise. The choice of \( k \) successes from \( M \) and the choice of \( n-k \) failures from \( N-M \) are calculated through these coefficients, contributing to the formula that determines the probability of a certain number of successes in a random sample.
Probability Calculation
Probability calculation is the mathematical process used to determine the likelihood of a particular outcome within a set of possible outcomes. When calculating probabilities, the outcome we consider successful is juxtaposed against all possible outcomes. In the context of hypergeometric distribution, we find the probability of a discrete number of successes in a sample.

This calculation involves using specialized formulas, such as the one seen in our hypergeometric distribution example, which is given by \( P(X=k) = \frac{\binom{M}{k}\binom{N-M}{n-k}}{\binom{N}{n}} \). The values incorporated into the formula, specifically \( M \) successes out of \( N \) total possible, and their respective binomial coefficients, frame a detailed picture of all probable selections.

Notably, the calculation of probabilities must respect the conditions of the distribution. For the hypergeometric distribution, it means considering the absence of replacement – once an item is selected from the population, it is not replaced, thus altering the probabilities of subsequent drawings. Additionally, probability calculations will often involve summing individual probabilities to get cumulative probabilities, as was done for our exercise to find the probability of observing at least two successes.
Discrete Random Variables
A discrete random variable is a type of random variable that can take on a finite or countably infinite set of distinct values. Such variables are the cornerstone of discrete probability distributions, notably the binomial and hypergeometric distributions.

Discrete random variables contrast with continuous random variables, which can take on an infinite number of values within a range. With discrete variables, we can enumerate each possible outcome and assign a probability to each. In our exercise, the number of successes \( X \) in the hypergeometric scenario is a discrete random variable because it can only take on a specific number of values (in this case, 0 through 5).

When working with discrete random variables, particularly in hypergeometric distributions, the sum of the probabilities for all possible outcomes always equals 1. This is due to the fact that one of the possible outcomes must occur. Being discrete allows us to calculate and visualize each probability separately, and if necessary, sum certain probabilities to gain insights into specific intervals or thresholds within the range of possible outcomes.

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Most popular questions from this chapter

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