/*! 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 Recall that by definition, a dis... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 22

Recall that by definition, a discrete sample space may contain a countably infinite number of outcomes. This exercise gives an example of such a countably infinite sample space. Suppose we flip a fair coin until it comes up heads. Of course, there is no way to know in advance how many flips will be required. Design a sample space and a probability density to model this situation. Prove that the probability density you define is legitimate.

Short Answer

Expert verified
The sample space is \( \{ H, TH, TTH, \ldots \} \) and the probability density is \( P(X=n) = \left( \frac{1}{2} \right)^n \) for \( n \geq 1 \), which sums to 1.

Step by step solution

01

Define the Sample Space

The sample space \( S \) for this scenario is the set of all possible sequences of coin flips where the first head appears. This can be represented as \( S = \{ H, TH, TTH, TTTH, \ldots \} \). Each outcome corresponds to a sequence where a tail (T) is followed by another tail, and so on, until the first head (H) appears. Thus, each outcome is identified by the number of trials it took to achieve the first head.
02

Define the Probability Density

The probability of achieving the first head on the \( n \)-th flip is \( P(X = n) \). For a fair coin, the probability of a tail followed by the first head at the \( n \)-th flip is \( \left( \frac{1}{2} \right)^{n-1} \times \frac{1}{2} = \left( \frac{1}{2} \right)^n \). Hence, \( P(X = n) = \left( \frac{1}{2} \right)^n \) for \( n = 1, 2, 3, \ldots \).
03

Verify the Probability Distribution

To verify this is a legitimate probability distribution, we must show that the probabilities sum to 1. We calculate the sum: \[ \sum_{n=1}^{\infty} P(X = n) = \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \] This is an infinite geometric series with first term \( a = \frac{1}{2} \) and ratio \( r = \frac{1}{2} \). The sum of a geometric series is given by \( \frac{a}{1-r} \), which in this case is: \[ \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 \] Thus, the probabilities sum to 1, confirming the distribution is valid.

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.

Probability Density
When dealing with discrete sample spaces, we often need to define a probability density to assign probabilities to each possible outcome. In the context of flipping a fair coin until a head appears, the probability density function helps us determine the likelihood of achieving this outcome on a specific flip.
For the given problem, the probability of getting the first head on the \( n \)-th flip is represented by \( P(X = n) = \left( \frac{1}{2} \right)^n \). This formula is derived from multiplying the probability of getting consecutive tails (\( \frac{1}{2} \) for each tail) followed by a head (\( \frac{1}{2} \)).
Such a function must satisfy two conditions:
  • Each probability value must be between 0 and 1.
  • The sum of all probabilities must equal 1.
This ensures that the probability distribution is accurate and complete.
Geometric Series
A geometric series is a series of terms that have a constant ratio between successive terms. In our problem, this series represents the total probability of getting heads at any time during the sequence of coin flips.
The probability for the first head to occur is expressed as an infinite geometric series: \( \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots \). This series has a first term \( a = \frac{1}{2} \) and a common ratio \( r = \frac{1}{2} \).
The formula for the sum of a geometric series \( S \) is given by \( \frac{a}{1-r} \). In this case, substituting the values, we find \( \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 \). This confirms that the series sums to 1, validating the entire probability distribution.
Fair Coin
A fair coin is a theoretical concept where the coin has an equal likelihood of landing on either heads or tails. This means the probability of getting a head is \( \frac{1}{2} \), as is the probability of getting a tail, making it unbiased.
In our exercise, assuming a fair coin model allows us to predict each flip's outcome with this simple probability, simplifying the probability density calculations.
With the sequence of events (like getting multiple tails before a head), the fair nature of the coin leads to multiplying \( \frac{1}{2} \) for every additional flip. Understanding a fair coin's principle is crucial because it forms the foundation of calculating probabilities in sequential experiments like the coin flipping exercise in question.

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

Two nickels and a dime are shaken together and thrown. We are allowed to keep the coins that turn up heads. We choose a sample space \(\Omega=\\{0,5,10,15,20\\},\) the outcomes of which correspond to the amounts that we can keep. For each of the following situations, cither describe the situation as an event in \(\Omega\) by listing the elements in the appropriate subset of \(\Omega\) or state that the situation cannot be described as an event in this particular sample space: (a) No heads. (b) All heads. (c) Exactly one coin turns up heads. (d) Exactly one of the nickels tums up heads. (e) The dime turns up heads.

The probability density function for the random variable \(X\) defined to be the number of cars owned by a randomly selected family in Millinocket is given as $$\begin{array}{l|c|c|c|c|c}x & 0 & 1 & 2 & 3 & 4 \\\\\hline p(X=x) & 0.08 & 0.15 & 0.45 & 0.27 & 0.05\end{array}$$ Compute the variance and standard deviation of \(X\).

Two nickels and a dime are shaken together and thrown. All the coins are fair. We are allowed to keep the coins that turn up heads. Give two sample spaces together with probability density functions that reasonably describe this situation. Explain your answer.

Suppose that \(\sum_{l=1}^{n} a_{i}=\sum_{j=1}^{m} b_{j}=1\) where \(0 \leq a_{i}, b_{j} \leq 1 .\) Use the Product of Sums Principle to prove that \(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i} \cdot b_{j}=1 .\) Does the result hold if some of the \(a_{i}\) and \(b_{j}\) can be less than zero and greater than one?

Two manufacturing companies \(M_{1}\) and \(M_{2}\) produce a certain unit that is used in an assembly plant. Company \(M_{1}\) is larger than \(M_{2}\), and it supplies the plant with twice as many units per day as \(M_{2}\) does. \(M_{1}\) also produces more defects than \(M_{2}\). Because of past experience with these suppliers, it is felt that \(10 \%\) of \(M_{1}\) 's units have some defect, whereas only \(5 \%\) of \(M_{2}\) 's units are defective. Now, suppose that a unit is selected at random from a bin in the assembly plant. (a) What is the probability that the unit was supplied by company \(M_{1} ?\) (b) What is the probability that the unit is defective? (c) What is the probability that the unit was supplied by \(M_{1}\) if the unit is defective?
See all solutions

Recommended explanations on Computer Science Textbooks

Functional Programming

Read Explanation

Computer Systems

Read Explanation

Data Structures

Read Explanation

Theory of Computation

Read Explanation

Blockchain Technology

Read Explanation

Computer Network

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.