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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 probabilities are as follows: - No successes: 0 - At least two successes: 246/252 - Exactly two successes: 60/252

Step by step solution

01

Set up the hypergeometric distribution formula for 0 successes

We want to find the probability of observing no successes (k=0) in a sample of n=5 items selected from N=10, where 6 items are considered successes (K=6). Set up the formula: \(P(X=0)=\frac{{6 \choose 0}{4 \choose 5}}{{10 \choose 5}}\)
02

Calculate the probability

Calculate the combinations and probabilities: \({6 \choose 0}=1\) \({4 \choose 5}=0\) \({10 \choose 5}=252\) \(P(X=0)=\frac{1 \cdot 0}{252}=0\) The probability of observing no successes is 0. #b. Find the probability of observing at least two successes.#
03

Set up the hypergeometric distribution formula for at least 2 successes

We want to find the probability of observing at least two successes, which means 2, 3, 4, or 5 successes. We will have to calculate the probability for each of these cases and then sum up the probabilities. \(P(X \geq 2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)\)
04

Calculate the probabilities and sum them up

Calculate the probabilities for each case and sum them up: \(P(X=2)=\frac{{6 \choose 2}{4 \choose 3}}{{10 \choose 5}}=\frac{15 \cdot 4}{252}=\frac{60}{252}\) \(P(X=3)=\frac{{6 \choose 3}{4 \choose 2}}{{10 \choose 5}}=\frac{20 \cdot 6}{252}=\frac{120}{252}\) \(P(X=4)=\frac{{6 \choose 4}{4 \choose 1}}{{10 \choose 5}}=\frac{15 \cdot 4}{252}=\frac{60}{252}\) \(P(X=5)=\frac{{6 \choose 5}{4 \choose 0}}{{10 \choose 5}}=\frac{6 \cdot 1}{252}=\frac{6}{252}\) \(P(X \geq 2)=\frac{60}{252}+\frac{120}{252}+\frac{60}{252}+\frac{6}{252}=\frac{246}{252}\) The probability of observing at least two successes is \(\frac{246}{252}\). #c. Find the probability of observing exactly two successes#
05

Use the hypergeometric distribution formula for 2 successes

We want to find the probability of observing exactly two successes (k=2) in a sample of n=5 items selected from N=10, where 6 items are considered successes (K=6). We've already calculated this probability in the previous part: P(X=2)
06

Recall the probability

The probability of observing exactly two successes is \(P(X=2)=\frac{60}{252}\).

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Key Concepts

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

Probability of Success
The probability of success in the context of hypergeometric distribution is about figuring out the likelihood of achieving a certain number of successes in a sample taken from a larger population. Since hypergeometric distribution deals with scenarios of sampling without replacement, the probability of success differs from that in cases where events are independent, such as in the binomial distribution. This means each success or failure changes the odds for next draws due to the diminishing pool of available outcomes. To calculate such probabilities, we leverage combinatorial analysis, specifically combinations, to arrange the possible successful outcomes over the entire set. For example, if we want to calculate the probability of exactly two successes when drawing five items from ten, where six are successes, we analyze from various angles:
  • How many ways we can choose two successful items from a total of six.
  • How many ways we can choose the remaining items to form the total sample from the non-successful set.
This helps in understanding how to manipulate odds in finite populations when outcomes are mutually dependent due to the nature of the sampling.
Combinatorial Analysis
Combinatorial analysis is integral to solving hypergeometric distribution problems. It involves determining the number of different ways to select a group of items from a larger pool, which is crucial when dealing with scenarios of sampling without replacement. In these cases, we use the combination formula \({n \choose k}\) to calculate the ways to choose \(k\) successes from \(n\) available successes, as well as the failures. Combination formulas work because we are interested in grouping items, and the order does not matter.
  • The formula \({n \choose k}\) = \(\frac{n!}{k!(n-k)!}\) where \'!\' represents factorial, meaning the product of all positive integers up to that number.
  • For instance, if six items are considered successes out of ten, \({6 \choose 2}\) shows how to select two success items from these.
Applying this idea of combinations, we can precisely determine probabilities in a hypergeometric framework. Each calculation reflects how components of the selection process furnish the complete picture for scenario-based probabilities, thus explaining why probabilities are connected to specific outcomes and total sampling.
Sampling Without Replacement
Sampling without replacement is a key concept in understanding hypergeometric distribution because it affects the probability calculation by altering the sample space with each draw. Unlike sampling with replacement, where the odds remain constant because you 'put back' and redraw, sampling without replacement changes the composition of the pool you are sampling from after every draw. This concept significantly influences the process and results of finding probabilities since it involves selecting from a shrinking pool. When an item is chosen, it isn't replaced, leading to a dynamic probability calculation. For example, if choosing from a group of ten items, once an item is drawn, only nine are left for subsequent draws. This directly impacts the combinations that can be formed, sustaining its relevance in problems like calculating the chances of successes in non-replenished sample sets:
  • As we progress through the sampling, both the number of potential successes and failures diminish.
  • The likelihood of further successes generally decreases with every success removed unless the pool favors more successes.
Thus, understanding how this sampling affects individual odds is vital for accurately solving probability problems where items aren’t replaced after each draw.

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Most popular questions from this chapter

The National Hockey League (NHL) has \(80 \%\) of its players born outside the United States, and of those born outside the United States, \(50 \%\) are born in Canada. \(^{2}\) Suppose that \(n=12\) NHL players were selected at random. Let \(x\) be the number of players in the sample who were born outside of the United States so that \(p=.8\). Find the following probabilities: a. At least five or more of the sampled players were born outside the United States. b. Exactly seven of the players were born outside the United States. c. Fewer than six were born outside the United States.

How do you survive when there's no time to eat-fast food, no food, a protein bar, candy? A Snapshot in USA Today indicates that \(36 \%\) of women aged \(25-55\) say that, when they are too busy to eat, they get fast food from a drive-thru. \({ }^{13}\) A random sample of 100 women aged \(25-55\) is selected. a. What is the average number of women who say they eat fast food when they're too busy to eat? b. What is the standard deviation for the number of women who say they eat fast food when they're too busy to eat? c. If 49 of the women in the sample said they eat fast food when they're too busy to eat, would this be an unusual occurrence? Explain.

Suppose that \(10 \%\) of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected and checked for whitefly. a. What is the average number of fields sampled that are infested with whitefly? b. Within what limits would you expect to find the number of infested fields, with probability approximately \(95 \% ?\) c. What might you conclude if you found that \(x=25\) fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.

A Snapshot in USA Today shows that \(60 \%\) of consumers say they have become more conservative spenders. \({ }^{12}\) When asked "What would you do first if you won \(\$ 1\) million tomorrow?" the answers had to do with somewhat conservative measures like "hire a financial advisor," or "pay off my credit card," or "pay off my mortgage.' Suppose a random sample of \(n=15\) consumers is selected and the number \(x\) of those who say they have become conservative spenders recorded. a. What is the probability that more than six consumers say they have become conservative spenders? b. What is the probability that fewer than five of those sampled have become conservative spenders? c. What is the probability that exactly nine of those sampled are now conservative spenders.

Americans are really getting away while on vacation. In fact, among small business owners, more than half \((51 \%)\) say they check in with the office at least once a day while on vacation; only \(27 \%\) say they cut the cord completely. \({ }^{7}\) If 20 small business owners are randomly selected, and we assume that exactly half check in with the office at least once a day, then \(n=20\) and \(p=.5 .\) Find the following probabilities. a. Exactly 16 say that they check in with the office at least once a day while on vacation. b. Between 15 and 18 (inclusive) say they check in with the office at least once a day while on vacation. c. Five or fewer say that they check in with the office at least once a day while on vacation. Would this be an unlikely occurrence?

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