/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Explain what is meant by the lon... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Explain what is meant by the long-run relative frequency definition of probability.

Short Answer

Expert verified
The long-run relative frequency definition of probability is the value that the ratio of favorable outcomes to total trials approaches as the number of trials becomes very large.

Step by step solution

01

Introduce Probability Concepts

Probability measures how likely an event is to occur. It's expressed as a number between 0 (impossible event) and 1 (certain event).
02

Define Long-Run Relative Frequency

In probability, the long-run relative frequency is defined as the value that the relative frequency of an event approaches when the number of trials becomes very large. It provides an empirical estimate of the probability based on repeated experimentation.
03

Explain the Theory Behind It

The long-run relative frequency definition suggests that if you repeat an experiment many times, the probability of an event can be approximated by the ratio of the number of times the event occurs to the total number of trials. In other words, as the number of trials approaches infinity, this ratio approaches the true probability of the event.
04

Provide an Example

Consider flipping a fair coin. In theory, the probability of getting heads is 0.5. If you flip the coin many times, the fraction of heads will approach 0.5 as the number of trials increases, illustrating the long-run relative frequency.

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Key Concepts

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

Long-Run Relative Frequency
Probability serves as a foundation for understanding the likelihood of various outcomes. One way to grasp probability is through the concept of long-run relative frequency. This idea revolves around observing the frequency of an event as a ratio that tends to stabilize as the number of trials increases. Think of it like watching the results of rolling a six-sided die. If you roll the die a few times, you might not see every number equally. But as you continue rolling it, say a hundred or even a thousand times, the frequency of each number appearing will get closer to one-sixth. This is because, in the long run, the die should behave fairly. The long-run relative frequency provides a practical way to estimate the likelihood of results from repeated experimental trials.
Empirical Estimate
Before diving into the specifics of probability, it's necessary to understand what makes an estimation empirical. Empirical estimates are grounded in observation and experimentation rather than theory alone. They manifest through practical reasoning when conducting experiments over time. In the context of probability, empirical estimates come from repeating an experiment numerous times to see how often a particular outcome occurs. Consider observing the weather: over many days, you could empirically estimate the probability of rain on any given day by counting the number of rainy days and dividing this by the total number of days. This grounded approach allows for a robust understanding of the likelihood, serving as a key element in determining probabilities through real-life data.
Experimentation in Probability
Probability thrives on experimentation, which offers a tangible way to approach and understand randomness. Through experimentation, such as repeated trials or simulations, we can observe patterns and calculate probabilities. These experiments often start with a simple setup, like flipping a coin. Through observing and recording the outcome, repetition builds a clearer picture of likelihood and randomness. The more experiments carried out, the more an empirical estimate of probability becomes refined. Experimentation helps solidify abstract concepts into everyday understanding, enhancing practical reasoning through hands-on activities. This approach is essential for anyone learning about probability, as it connects theoretical probabilities to real-world scenarios, equipping learners with tools to navigate uncertainty effectively.

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Most popular questions from this chapter

Airplane safety has been improving over the years. From 2000 to \(2010,\) the average number of global airline deaths per year was over 1000 , even when excluding the nearly 3000 deaths in the United States on September 11,2001 . The number of global airline deaths declined in 2011 , again in \(2012,\) and then hit a low of only 265 in \(2013 .\) In \(2013,\) there were a total of 825 million passengers globally. Sources: en.wikipedia.org and www.transtats.bts.gov1. a. Can you consider the 2013 data as a long run or short run of trials? Explain. b. Estimate the probability of dying on a flight in \(2013 .\) (Note, the probability of dying from a 1000 -mile automobile trip is about 1 in 42,000 by contrast.) c. Raul is considering flying on an airplane. He noticed that over the past two months, there have been no fatal airplane crashes around the world. This raises his concern about flying because the airlines are "due for an accident." Comment on his reasoning.

The powerrank.com website (http:// thepowerrank.com/2014/06/06/world- cup-2014-winprobabilities-from-the-power-rank/) listed the probability of each team to win the 2014 World Cup in soccer as follows: 1\. Brazil, \(35.9 \%\). 2\. Argentina, \(10.0 \%\). 3\. Spain, \(8.9 \%\). 4\. Germany, \(7.4 \%\). 5\. Netherlands, \(5.7 \%\). 6\. Portugal, \(3.9 \%\). 7\. France, \(3.4 \%\). 8\. England, \(2.8 \%\). 9\. Uruguay, \(2.5 \%\). 10\. Mexico, \(2.5 \%\). 11\. Italy, \(2.3 \%\). 12\. Ivory Coast, \(2.0 \%\), 13\. Colombia, \(1.5 \%\). 14\. Russia, \(1.5 \%\). 15\. United States, \(1.1 \%\). 16\. Chile, \(1.0 \%\). 17\. Croatia, \(0.9 \%\) 18\. Ecuador, \(0.8 \%\). 19\. Nigeria, \(0.8 \%\). 20\. Switzerland, \(0.7 \%\). 21\. Greece, \(0.6 \%\) 22\. \(\operatorname{Iran}, 0.6 \%\). 23\. Japan, \(0.6 \%\). 24\. Ghana, \(0.6 \%\). 25\. Belgium, \(0.4 \%\). 26\. Honduras, \(0.3 \%\). 27\. South Korea, \(0.3 \%\). 28\. Bosnia-Herzegovina, \(0.3 \%\). 29\. Costa Rica, \(0.3 \%\). 30\. Cameroon, \(0.2 \%\). 31\. Australia, \(0.2 \%\). 32\. Algeria, \(0.1 \%\). a. Interpret Brazil's probability of \(35.9 \%,\) which was based on computer simulations of the tournament. Is it a relative frequency or a subjective interpretation of probability? b. Germany would emerge as the actual winner of the 2014 World Cup. Does this indicate that the \(7.4 \%\) chance of Germany winning, which was calculated before the tournament, should have been \(100 \%\) instead?

Your friend is interested in estimating the proportion of people who would vote for his project in a local contest. He selects a large sample among his many friends and claims that, with such a large sample, he does not need to worry about the method of selecting the sample. What is wrong in this reasoning? Explain.

In the opening scene of Tom Stoppard's play Rosencrantz and Guildenstern Are Dead, about two Elizabethan contemporaries of Hamlet, Guildenstern flips a coin 91 times and gets a head each time. Suppose the coin was balanced. a. Specify the sample space for 91 coin flips, such that each outcome in the sample space is equally likely. How many outcomes are in the sample space? b. Show Guildenstern's outcome for this sample space. Show the outcome in which only the second flip is a tail. c. What's the probability of the event of getting a head 91 times in a row? d. What's the probability of at least one tail in the 91 flips? e. State the probability model on which your solutions in parts \(\mathrm{c}\) and \(\mathrm{d}\) are based.

Online sections For a course with two sections, let \(\mathrm{A}\) denote \\{first section is online\\}, let \(\mathrm{B}\) denote (at least one sec- tion is online\\}, and let \(C\) denote (both sections are online\\}. Suppose \(\mathrm{P}\) (a section is online) \(=1 / 2\) and that the sections are independent. a. Find \(\mathrm{P}(\mathrm{C} \mid \mathrm{A})\) and \(\mathrm{P}(\mathrm{C} \mid \mathrm{B})\). b. Are \(A\) and \(C\) independent events? Explain why or why not. c. Describe what makes \(\mathrm{P}(\mathrm{C} \mid \mathrm{A})\) and \(\mathrm{P}(\mathrm{C} \mid \mathrm{B})\) different from each other.

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