/*! 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 22 You are given four coins: one ha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 22

You are given four coins: one has two heads, one has two tails, and the other two are normal. You choose a coin at random and toss it. The result is tails. What is the probability that the opposite face is heads?

Short Answer

Expert verified
The probability is \( \frac{1}{2} \).

Step by step solution

01

Define the sample space

First, define the possible outcomes when you choose a coin: you could select the coin with two heads (HH), the one with two tails (TT), or either of the two normal coins (HT and HT again). Since there are two normal coins, we treat them as two unique selections.
02

Determine the probability of selecting each type of coin

The probability of selecting the coin with two heads (HH) is \( \frac{1}{4} \), the probability of selecting the coin with two tails (TT) is \( \frac{1}{4} \), and the probability of selecting each of the normal coins (HT) is \( \frac{1}{4} \) each.
03

Identify the probabilities of getting tails on the first toss

Now, calculate the probability of getting tails on the first toss from each type of coin:- Coin HH: Probability is 0 (impossible to get tails)- Coin TT: Probability is 1 (always get tails)- Each Coin HT: Probability is \( \frac{1}{2} \) (one head, one tail, equally likely).
04

Apply Bayes' Theorem

We need to find the probability that a normal coin was chosen, given that the result of the toss was tails. Use Bayes' Theorem:\[ P(A|B) = \frac{P(B|A)P(A)}{P(B)} \]Let A be the event that a normal coin was chosen, and B be the event that tails appears.- \( P(A) \) (probability of choosing a normal coin) = \( \frac{1}{2} \)- \( P(B|A) \) (probability of getting tails given normal coin) = \( \frac{1}{2} \)- \( P(B) \) (probability of getting tails):\[ P(B) = P(B|HH)P(HH) + P(B|TT)P(TT) + P(B|HT)P(HT) \]\[ P(B) = 0 \times \frac{1}{4} + 1 \times \frac{1}{4} + \frac{1}{2} \times \frac{2}{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]
05

Calculate the probability

Using the computed values, we find:\[ P(A|B) = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{1}{2}} = \frac{1}{2} \]Thus, the probability that the opposite face is heads is \( \frac{1}{2} \).

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 91影视!

Key Concepts

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

Bayes' Theorem
Bayes' Theorem is a powerful mathematical formula used to find the probability of an event based on prior knowledge of related events. It is particularly handy in dealing with problems involving conditional probabilities. In our coin problem, we want to find the probability that a normal coin was chosen given that the result of the toss was tails. This involves using existing information, like the probability of choosing each type of coin and the probability of getting tails with them.
  • Bayes' Theorem Formula: \( P(A|B) = \frac{P(B|A)P(A)}{P(B)} \)
In the formula:
  • \(P(A|B)\) is the probability of event A occurring given event B has occurred.
  • \(P(B|A)\) is how likely event B is to occur given A has occurred.
  • \(P(A)\) is the initial probability of event A.
  • \(P(B)\) is the total probability of event B.
By applying this fundamental theorem, we make sense of how the likelihood of picking a normal coin can be calculated after landing on tails.
Sample Space
The concept of a sample space is central to probability theory. It represents all possible outcomes of a probabilistic experiment. In simpler terms, it is the set of everything that can happen when we perform a random trial. When dealing with coin problems, the sample space is straightforward. For our problem, the sample space consists of the coins that could potentially be selected.
In this scenario:
  • The coins include: two-headed (HH), two-tailed (TT), and two normal coins that have one head and one tail (HT, HT).
  • Each coin in the sample space has an equal chance of being selected, which is \(\frac{1}{4}\) for each coin due to equal likelihood.
  • Defining the sample space helps calculate probabilities systematically.
Having a clear sample space allows us to better understand our probability calculations and helps us implement methods like Bayes' Theorem effectively.
Coin Toss Problems
Coin toss problems are a classic example of experiments that involve calculating probability. They are often used because the outcomes are simple: heads or tails. However, problems can become complex when additional scenarios are involved, like having coins with different configurations. In our problem:
  • We have a mix of coins: one with two heads, one with two tails, and two normal coins.
  • The challenge lies in correctly identifying the probability of outcomes given these mixed conditions.
  • When a coin is tossed and results in tails, we solve for the probability of it being a normal coin using Bayes' Theorem.
These problems teach important concepts in probability, such as conditional probability and independent events, which are crucial in understanding more sophisticated topics in probability theory.

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

A small class has five boys and six girls. A group of four students are to be selected at random to interview a new school director. a) Find the probability that the group contains at least one boy. b) Find the probability that the majority of the group is girls. c) Given that the group contains at least one boy, what is the chance that the boys are in the majority?

The majority of email messages are now'spam? Choose a spam email message at random. Here is the distribution of topics: $$\begin{array}{|l|c|c|c|c|c|c|} \hline \text { Topic } & \text { Adult } & \text { Financial } & \text { Health } & \text { Leisure } & \text { Products } & \text { Scams } \\ \hline \text { Probability } & 0.165 & 0.142 & 0.075 & 0.081 & 0.209 & 0.145 \\\ \hline \end{array}$$ a) What is the probability of choosing a spam message that does not concern these topics? Parents are usually concerned with spam messages with 'adult'content and scams. b) What is the probability that a randomly chosen spam email falls into one of the other categories?

A construction company is bidding on three projects: \(B_{1}, B_{2}\) and \(B_{3}\). From previous experience they have the following probabilities of winning the bids: \(P\left(B_{1}\right)=0.22, P\left(B_{2}\right)=0.25\) and \(P\left(B_{3}\right)=0.28 .\) Winning the bids is not independent one from another. The joint probabilities are given below. $$\begin{array}{|c|c|c|c|} \hline & B_{1} & B_{2} & B_{3} \\ \hline B_{1} & & 0.11 & 0.05 \\ \hline B_{2} & 0.11 & & 0.07 \\ \hline B_{3} & 0.05 & 0.07 & \\ \hline \end{array}$$ Also, \(P\left(B_{1} \cap B_{2} \cap B_{3}\right)=0.01 .\) Find the following probabilities: a) \(P\left(B_{1} \cup B_{2}\right)\) b) \(P\left(B^{\prime}_{1} \cap B_{2}^{\prime}\right)\) c) \(P\left(B^{\prime}, \cap B_{2}^{\prime}\right) \cup B_{3}\) d) \(P\left(B^{\prime}_{1} \cap B_{2}^{\prime} \cap B_{3}\right)\) e) \(P\left(B_{2} \cap B_{3} | B_{1}\right)\) \(f) P\left(B_{2} \cup B_{3} | B_{1}\right)\)

When answering a question on a multiple choice test, a student is given 5 choices, one of which is correct. The test is so designed that the choices are very close and the probability of getting the correct answer, when you know the material, is \(0.6 .\) In a class where \(70 \%\) of the students are well prepared, a randomly chosen student answers the question correctly. What is the probability that the student really knew the material?

During a dinner party, Magda plans on opening six bottles of wine. Her supply includes 8 French, 10 Australian and 12 Italian wines. She sends her sister Mara to choose the bottles. Mara has no knowledge of the wine types and picks the bottles at random. a) What is the probability that two of each type get selected? b) What is the probability that all served bottles are of the same type? c) What is the probability of serving only Italian and French wines?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation
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.