/*! 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 93 A study has shown that seven out... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A study has shown that seven out of ten people will say "heads" if asked to call a coin toss. Given that the coin is fair, though, a head occurs, on the average, only five times out of ten. Does it follow that you have the advantage if you let the other person call the toss? Explain.

Short Answer

Expert verified
No, it does not follow that you have the advantage if you let the other person call the toss. The probability of getting 'heads' or 'tails' is always equal in a fair toss, regardless of who calls it or what they call.

Step by step solution

01

Understanding the Problem Statement

From the problem statement, it is understood that seven out of ten people will say 'heads' if asked to call a coin toss. But we know that the probability of getting a 'head' in a fair coin toss is 0.5 or five times out of ten.
02

Analyzing the Situation

We need to understand that the choice of a person calling 'heads' or 'tails' and the actual outcome of a coin toss are two independent events. The outcome of the coin toss does not depend at all on what the person chooses.
03

Concluding the Analysis

Given these considerations, it becomes clear that even if the majority of people choose 'heads', this does not affect the actual probability of the coin landing on 'heads'. Thus, the statement 'you have the advantage if you let the other person call the toss' is incorrect because the chance is always equal (0.5), regardless of who calls the toss.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Independent Events
In probability theory, independent events are two or more events where the occurrence of one event does not affect the occurrence of another. This is an essential concept when evaluating scenarios such as a coin toss. When we say two events are independent, it means that the outcome of the first event has no impact on the outcome of the second event.

For example, consider a classical fair coin toss. Whether you flip the coin once or twice, getting heads on the first flip does not change the likelihood of getting heads on the second flip. Each flip is an independent event. Therefore, the outcomes do not influence each other, and this independence plays a crucial role in predicting probabilities in various situations.

In the context of allowing someone to guess the outcome beforehand, such as calling 'heads', their choice does not influence the actual result of the coin toss. The guess and the actual result are independent. So even if every person says 'heads', it doesn't change the odds of the coin actually landing on heads.
Fair Coin Toss
A fair coin toss is a classic example used to illustrate fundamental probability concepts. A coin is called fair if it has an equal chance of landing on either side, heads or tails, each time it is tossed. This means the probability of getting heads (or tails) in a single toss is 0.5. This simple and clear-cut scenario is a great tool for introducing probability theory.

The concept of fairness is crucial here. With a fair coin, the physical properties ensure that over a very large number of tosses, heads and tails will come up about equally often. This makes it an ideal tool for games of chance, as it ensures no bias towards one outcome.

Given the fairness, whether you toss the coin or someone else does, or whether you are the one calling it or the other person, the probability remains constant. There is no strategy or advantage to exploit, as every outcome is equally likely.
Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random phenomena. It provides a way to quantify uncertainty and predict the likelihood of different potential outcomes. In simplest terms, probability helps us answer the question: "What are the chances?"

In any probabilistic experiment, such as a coin toss, each possible outcome is assigned a probability. The sum of probabilities for all possible outcomes always equals 1. For a fair coin, the probabilities are distributed equally between heads and tails, hence each has a probability of 0.5.

Probability theory uses models such as the probability space, consisting of a sample space of all possible outcomes, events (collections of outcomes), and a probability measure that assigns probabilities to events. Understanding these concepts helps us make reasoned predictions based on known odds, and recognize situations where randomness plays a governing role. Whether dealing with coins, dice, or real-life scenarios, probability theory guides us in making informed decisions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In an upstate congressional race, the incumbent Republican \((R)\) is running against a field of three Democrats \(\left(D_{1}, D_{2}\right.\), and \(\left.D_{3}\right)\) seeking the nomination. Political pundits estimate that the probabilities of \(D_{1}, D_{2}\), or \(D_{3}\) winning the primary are \(0.35,0.40\), and \(0.25\), respectively. Furthermore, results from a variety of polls are suggesting that \(R\) would have a \(40 \%\) chance of defeating \(D_{1}\) in the general election, a \(35 \%\) chance of defeating \(D_{2}\), and a \(60 \%\) chance of defeating \(D_{3}\). Assuming all these estimates to be accurate, what are the chances that the Republican will retain his seat?

Building permits were issued last year to three contractors starting up a new subdivision: Tara Construction built two houses; Westview, three houses; and Hearthstone, six houses. Tara's houses have a \(60 \%\) probability of developing leaky basements; homes built by Westview and Hearthstone have that same problem \(50 \%\) of the time and \(40 \%\) of the time, respectively. Yesterday, the Better Business Bureau received a complaint from one of the new homeowners that his basement is leaking. Who is most likely to have been the contractor?

In a roll of a pair of fair dice (one red and one green), let \(A\) be the event of an odd number on the red die, let \(B\) be the event of an odd number on the green die, and let \(C\) be the event that the sum is odd. Show that any pair of these events is independent but that \(A, B\), and \(C\) are not mutually independent.

A fair coin is tossed four times. What is the probability that the number of heads appearing on the first two tosses is equal to the number of heads appearing on the second two tosses?

Let \(A_{1}, A_{2}, \ldots, A_{k}\) be any set of events defined on a sample space \(S\). What outcomes belong to the event $$ \left(A_{1} \cup A_{2} \cup \cdots \cup A_{k}\right) \cup\left(A_{1}^{C} \cap A_{2}^{C} \cap \cdots \cap A_{k}^{C}\right) $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.