/*! 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 Toss a fair coin twice. Let \(X\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 1

Toss a fair coin twice. Let \(X\) be the random variable that counts the number of tails in each outcome. Find the probability mass function describing the distribution of \(X\).

Short Answer

Expert verified
The PMF is \( P(X=x) = \{\frac{1}{4}, \frac{1}{2}, \frac{1}{4}\} \) for \( x=0, 1, 2 \).

Step by step solution

01

List All Possible Outcomes

A fair coin has two possible outcomes when tossed: heads (H) or tails (T). For two tosses, list all possible combinations: HH, HT, TH, TT. These represent the sample space S of this probability experiment.
02

Define the Random Variable X

Let the random variable \(X\) represent the number of tails in the outcome of the two coin tosses. Assign values to \(X\) based on the outcomes: X = 0 (HH), X = 1 (HT, TH), X = 2 (TT).
03

Calculate Probabilities for X = 0

Count the outcomes that result in 0 tails, which is the outcome HH. There is 1 outcome, so the probability \( P(X=0) = \frac{1}{4} \).
04

Calculate Probabilities for X = 1

Count the outcomes that result in 1 tail, which are HT and TH. There are 2 such outcomes, so the probability \( P(X=1) = \frac{2}{4} = \frac{1}{2} \).
05

Calculate Probabilities for X = 2

Count the outcomes that result in 2 tails, which is the outcome TT. There is 1 outcome, so the probability \( P(X=2) = \frac{1}{4} \).
06

Write the Probability Mass Function (PMF)

The probability mass function of \( X \) is given by: \[ P(X=x) = \begin{cases} \frac{1}{4}, & \text{if } x = 0, \ \frac{1}{2}, & \text{if } x = 1, \ \frac{1}{4}, & \text{if } x = 2. \end{cases} \] This function describes the distribution of the number of tails obtained.

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.

Understanding Random Variables
In probability and statistics, a random variable is a function that assigns a numerical value to each outcome in a sample space of a random experiment. Think of it as a way to "count" or "measure" certain outcomes in a mathematical way.

Here are the key aspects of random variables:
  • They represent real-world events numerically, providing a bridge between random experiments and numerical analysis.
  • Random variables can be discrete (having distinct values) or continuous (having a range of values).
  • In our exercise, the random variable \(X\) is discrete and counts the number of tails in the two coin tosses.
By defining \(X\), we can analyze the probability of different outcomes based on our coin toss scenario. This is a central concept for exploring probability distributions.
Understanding Coin Toss
A coin toss is one of the simplest examples of a random experiment. It's used frequently in probability theory because it provides a clear, binary outcome: heads (H) or tails (T). This characteristic makes it ideal for learning basic probability concepts.

Let's break down the coin toss experiment:
  • A single coin toss has two possible outcomes, making it a binomial experiment.
  • When tossing a fair coin, each outcome (H or T) has an equal probability of \(\frac{1}{2}\).
  • By considering two tosses, we expand the experimentation, forming pairs like HH, HT, TH, TT.
With multiple tosses, as seen in our exercise, you can see how probabilities combine to affect outcomes, building a more complex sample space.
Exploring Probability Distribution
Probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. Specifically, the probability mass function (PMF) gives the probability that a discrete random variable is exactly equal to some value.

Key points about probability distributions in this context:
  • For discrete random variables like \(X\), the PMF summarizes the likelihood of different outcomes within the sample space.
  • In our coin toss experiment, the PMF details probabilities for \(X = 0\), \(X = 1\), and \(X = 2\), based on observed outcomes.
  • The PMF shows that the probabilities add up to 1, which validates the total coverage of all possible events.
By understanding probability distributions, you can better analyze and predict the results of random experiments, like how likely certain numbers of tails appear in two coin tosses.

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

Suppose the number of phone calls arriving at a switchboard per hour is Poisson distributed with mean 7 calls per hour. Find the probability that no phone calls arrive during a certain hour.

Use the following facts: Cystic fibrosis is an inherited disorder that causes abnormally thick body secretions. About 1 in 2500 white babies in the United States has this disorder. About 3 in 100 children with cystic fibrosis develop diabetes mellitus, and about 1 in 5 females with cystic fibrosis is infertile. Find the probability that, in a group of 1000 children with cystic fibrosis, at least 25 will develop diabetes mellitus.

The following data represent the relative frequency distribution of clutch size in a sample of 42 mallards: $$ \begin{array}{cc} \hline \text { Clutch Size } & \text { Relative Frequency } \\ \hline 6 & 0.10 \\ 7 & 0.24 \\ 8 & 0.29 \\ 9 & 0.21 \\ 10 & 0.16 \\ \hline \end{array} $$ Calculate the sample mean and the sample variance.

An urn contains 1 black and 14 white balls. Balls are drawn at random, one at a time, until the black ball is selected. Each ball is replaced before the next ball is drawn. Find the probability that at least 20 draws are needed.

Toss a fair \(\operatorname{coin} 300\) times. (a) Use the central limit theorem and the histogram correction to find an approximation for the probability that the number of heads is between 140 and 160 . (b) Use Chebyshev's inequality to find an estimate for the event in (a), and compare your estimate with that in (a).
See all solutions

Recommended explanations on Biology Textbooks

Cellular Energetics

Read Explanation

Biological Processes

Read Explanation

Biology Experiments

Read Explanation

Cells

Read Explanation

Astrobiological Science

Read Explanation

Biological Organisms

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.