/*! 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 99 In a sequence of independent ide... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 99

In a sequence of independent identical trials with two possible outcomes on each trial, \(S\) and \(F\) and with \(P(S)=p,\) what is the probability that exactly \(y\) trials will occur before the rth success?

Short Answer

Expert verified
Probability is given by \(P(y-r\text{ failures before }r\text{ successes})=\binom{y-1}{r-1} p^r (1-p)^{y-r}\).

Step by step solution

01

Understanding the Question

In this problem, we have a sequence of trials where we either succeed (denoted by \(S\)) or fail (denoted by \(F\)). The probability of success in each trial is \(p\). We wish to find the probability of achieving the \(r\)-th success on the \(y\)-th trial.
02

Introducing the Negative Binomial Distribution

The problem is modeled using the negative binomial distribution. This distribution provides the probability that a fixed number of successes \(r\) will occur for the \(r\)-th time on the \(y\)-th trial. The negative binomial probability mass function (PMF) is given by:\[P(X = y) = \binom{y-1}{r-1} p^r (1-p)^{y-r}\] where \(X\) is the number of trials needed to achieve \(r\) successes.
03

Identifying Parameters and Applying the Formula

In this case, the parameters are: \(r\) is the number of successes we desire, \(y\) is the total number of trials, and \(p\) is the probability of success on each trial. Substitute these into the negative binomial PMF:\[P(X = y) = \binom{y-1}{r-1} p^r (1-p)^{y-r}\] This equation computes the probability of exactly \(r\) successes occurring on the \(y\)-th trial.
04

Conclusion

The probability that exactly \(y\) trials will occur before the \(r\)-th success is given by the negative binomial formula derived in the earlier steps. Substitute numerical values for \(r\), \(y\), and \(p\) as needed to calculate a specific result.

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 Theory
Probability theory is a key area of mathematics that studies the likelihood of occurrences. It applies to various events or outcomes, which can range from simple coin tosses to complex systems in nature.
In a probabilistic context, it helps in understanding situations where outcomes are uncertain. The core idea revolves around measuring the chances of various events using numbers.
  • Random Experiments: These are processes or actions with uncertain outcomes. For instance, rolling a die or flipping a coin.
  • Outcomes: These are the possible results of a random experiment. Each outcome has an associated probability.
  • Events: A collection of outcomes is considered an event. For example, rolling an even number on a die.
  • Probability: A measurement that quantifies the chance of an event occurring, typically ranging from 0 (impossible) to 1 (certain).
In probability theory, understanding the probability of getting the rth success in y trials is modeled by the negative binomial distribution. This distribution helps predict events when multiple trials are conducted.
Independent Trials
The concept of independent trials is crucial in probability theory, especially when discussing experiments with repeated actions.
Independent trials mean the outcome of one trial does not affect the outcome of another.
For instance, tossing a coin multiple times. Whether the first flip results in heads or tails does not change the odds of getting heads or tails in the following tosses.
Each coin flip is an independent trial.
  • Consistency of Probability: In independent trials, each trial has the same probability of success. If the probability of success is 0.5 for the first trial, it remains 0.5 for subsequent trials.
  • Applications: This concept applies to experiments where each trial or event is separate from others, such as drawing cards from a deck with replacement or rolling dice per trial.
  • Role in the Negative Binomial Distribution: Independent trials are foundational when calculating events like reaching the rth success at the yth attempt, using the negative binomial distribution.
Understanding independent trials allows for clearer insights into the probabilistic modeling of repeated actions and events.
Success and Failure
In probability, every trial of a random experiment can result in success or failure. This binary outcome plays a vital role in various probability distributions, including the negative binomial distribution.
A trial resulting in success is termed as 'Success' (S), while a trial not meeting the success criteria is termed as 'Failure' (F).
  • Success: This outcome signifies achieving the desired result. For example, if flipping heads in a coin toss is considered a success, getting a head means success.
  • Failure: This indicates not achieving the desired result in a trial. Continuing with the coin toss example, flipping a tail represents failure if heads are defined as the success condition.
  • Probability of Success: Denoted by the symbol \( p \), it refers to the likelihood of achieving success in a single trial.
  • Probability of Failure: Expressed as \( 1-p \), this is the likelihood of a trial resulting in failure.
  • Success and Failure in Negative Binomial Distribution: This distribution concerns finding the probability of a set number of successes within a sequence of independent trials, detailing how many trials are required for a certain number of successes, acknowledging both successes and failures along the way.
Ultimately, recognizing the binary nature of trial outcomes simplifies calculations in probabilistic models.

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 that \(Y\) is a binomial random variable based on \(n\) trials with success probability \(p\) and consider \(Y^{*}=n-Y\) a. Argue that for \(y^{*}=0,1, \ldots, n\) $$ P\left(Y^{*}=y^{*}\right)=P\left(n-Y=y^{*}\right)=P\left(Y=n-y^{*}\right) $$ b. Use the result from part (a) to show that $$ P\left(Y^{*}=y^{*}\right)=\left(\begin{array}{c} n \\ n-y^{*} \end{array}\right) p^{n-y^{*}} q^{y^{*}}=\left(\begin{array}{c} n \\ y^{*} \end{array}\right) q^{y^{*}} p^{n-y^{*}} $$ c. The result in part (b) implies that \(Y^{*}\) has a binomial distribution based on \(n\) trials and "success" probability \(p^{*}=q=1-p .\) Why is this result "obvious"?

About six months into George W. Bush's second term as president, a Gallup poll indicated that a near record (low) level of \(41 \%\) of adults expressed "a great deal" or "quite a lot" of confidence in the U.S. Supreme Court (http:// www.gallup.com/poll/content/default.aspx?ci=17011, June 2005). Suppose that you conducted your own telephone survey at that time and randomly called people and asked them to describe their level of confidence in the Supreme Court. Find the probability distribution for \(Y\), the number of calls until the first person is found who does not express "a great deal" or "quite a lot" of confidence in the U.S. Supreme Court.

Used photocopy machines are returned to the supplier, cleaned, and then sent back out on lease agreements. Major repairs are not made, however, and as a result, some customers receive malfunctioning machines. Among eight used photocopiers available today, three are malfunctioning. A customer wants to lease four machines immediately. To meet the customer's deadline, four of the eight machines are randomly selected and, without further checking, shipped to the customer. What is the probability that the customer receives a. no malfunctioning machines? b. at least one malfunctioning machine?

Find the distributions of the random variables that have each of the following moment-generating functions: a. \(m(t)=\left[(1 / 3) e^{t}+(2 / 3)\right]^{5}\) b. \(m(t)=\frac{e^{t}}{2-e^{t}}\) c. \(m(t)=e^{2\left(e^{t}-1\right)}\)

A shipment of 20 cameras includes 3 that are defective. What is the minimum number of cameras that must be selected if we require that \(P(\text { at least } 1 \text { defective) } \geq .8 ?\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

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.