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The negative binomial rv \(X\) was defined as the number of \(F\) 's preceding the \(r\) th \(S\). Let \(Y=\) the number of trials necessary to obtain the \(r\) th \(S\). In the same manner in which the pmf of \(X\) was derived, derive the pmf of \(Y\).

Short Answer

Expert verified
The pmf of \(Y\) is \( P(Y = n) = \binom{n-1}{r-1} p^r (1-p)^{n-r} \) for \(n \geq r\).

Step by step solution

01

Understanding the Problem

The task is to find the probability mass function (pmf) for the random variable \(Y\), which represents the number of trials needed to obtain the \(r\)th success \(S\) in a series of Bernoulli trials, with each trial having a probability \(p\) of success.
02

Relating \(Y\) and \(X\)

The random variable \(Y\) is related to \(X\) by the equation \(Y = X + r\). This is because \(X\) counts the number of failures \(F\) before the \(r\)th success, and we need an additional \(r\) successes to reach \(Y\).
03

Deriving the PMF Formula

Let the random variable \(Y\) take the value \(n\), then \(X = n-r\). The pmf of \(X\), the negative binomial, is: \[ P(X=n-r) = \binom{n-1}{r-1} p^r (1-p)^{n-r} \] This represents \(n-1\) trials including \(r\) successes and \(n-r\) failures.
04

Substituting and Final Formula

Since \(Y = X + r\), substitute \(n = X + r\) to write the pmf for \(Y\):\[ P(Y = n) = \binom{n-1}{r-1} p^r (1-p)^{n-r} \] This is valid for \(n = r, r+1, r+2, \ldots\), ensuring we have at least \(r\) trials for \(r\) successes.

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Key Concepts

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

Probability Mass Function
The probability mass function (PMF) is a fundamental concept in probability theory. It provides the probability that a discrete random variable is exactly equal to a specific value. In the case of the negative binomial distribution, it helps us calculate the likelihood of a particular number of trials until a predetermined number of successes occurs.

For the random variable \( Y \), which represents the number of trials needed to achieve the \( r \)-th success in a sequence of Bernoulli trials, the PMF gives us an explicit formula to calculate this probability. The formula is:
\[ P(Y = n) = \binom{n-1}{r-1} p^r (1-p)^{n-r} \] where:
  • \( n \) is the total number of trials.
  • \( r \) is the number of successful events needed.
  • \( p \) is the probability of success in each trial.
  • \( \binom{n-1}{r-1} \) is a binomial coefficient indicating the number of ways to choose \( r-1 \) successes in \( n-1 \) trials.
  • \( (1-p)^{n-r} \) accounts for the failures in the sequence.

Understanding the PMF allows us to determine the distribution of a random variable across different outcomes.
Bernoulli Trials
A Bernoulli trial is a basic random experiment often used in probability and statistics. Each trial has only two possible outcomes: success \( S \) or failure \( F \). The probability of success is denoted by \( p \), while the probability of failure is \( 1-p \). Bernoulli trials form the building blocks of many important probability distributions, including the negative binomial distribution.

In our scenario, achieving the \( r \)-th success means conducting several Bernoulli trials until we reach that goal. Each trial is independent, meaning the outcome of one trial doesn't affect the others. This independence is crucial for applying the formulas that describe these distributions, like the PMF we've discussed.

To solve problems involving Bernoulli trials, it's essential to clearly understand the definition and the settings of the trials:
  • Identify the probability \( p \) of success.
  • Determine how many successes \( r \) are needed.
  • Recognize that each trial contributes independently to the overall outcome.
By grasping these points, you'll effectively use Bernoulli trials to model real-world random processes.
Discrete Probability Distributions
Discrete probability distributions allocate probabilities to distinct and separate values. Unlike continuous distributions, which deal with intervals, discrete distributions focus on specific outcomes, making them ideal for situations involving countable events.

The negative binomial distribution, a type of discrete probability distribution, describes the number of trials required to achieve a fixed number of successes. It is particularly useful when results are skewed, such as needing more trials to achieve successes in scenarios with low success probabilities.

Within discrete distributions, probabilities are computed for distinct and countable numbers, reflecting specific measurements such as:
  • Number of trials until a certain number of successes.
  • Counting occurrences of a particular event in a sequence.
  • Discrete outcomes of games, surveys, or experiments.

These distributions give us a simple and structured way to deal with events that have specific outcomes. By understanding discrete distributions, we can better predict and analyze probabilistic events.

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Most popular questions from this chapter

In proof testing of circuit boards, the probability that any particular diode will fail is .01. Suppose a circuit board contains 200 diodes. a. How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail? b. What is the (approximate) probability that at least four diodes will fail on a randomly selected board? c. If five boards are shipped to a particular customer, how likely is it that at least four of them will work properly? (A board works properly only if all its diodes work.)

A test for the presence of a certain disease has probability \(.20\) of giving a false-positive reading (indicating that an individual has the disease when this is not the case) and probability . 10 of giving a false-negative result. Suppose that ten individuals are tested, five of whom have the disease and five of whom do not. Let \(X=\) the number of positive readings that result. a. Does \(X\) have a binomial distribution? Explain your reasoning. b. What is the probability that exactly three of the ten test results are positive?

Define a function \(p(x ; \lambda, \mu)\) by $$ p(x ; \lambda, \mu)=\left\\{\begin{array}{cl} \frac{1}{2} e^{-\lambda} \frac{\lambda^{x}}{x !}+\frac{1}{2} e^{-\mu \frac{\mu^{x}}{x !}} & x=0,1,2, \ldots \\ 0 & \text { otherwise } \end{array}\right. $$ a. Show that \(p(x ; \lambda, \mu)\) satisfies the two conditions necessary for specifying a pmf, [Note: If a firm employs two typists, one of whom makes typographical errors at the rate of \(\lambda\) per page and the other at rate \(\mu\) per page and they each do half the firm's typing, then \(p(x ; \lambda, \mu)\) is the pmf of \(X=\) the number of errors on a randomly chosen page.] b. If the first typist (rate \(\lambda\) ) types \(60 \%\) of all pages, what is the pmf of \(X\) of part (a)? c. What is \(E(X)\) for \(p(x ; \lambda, \mu)\) given by the displayed expression? d. What is \(\sigma^{2}\) for \(p(x ; \lambda, \mu)\) given by that expression?

The College Board reports that \(2 \%\) of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test. a. What is the probability that exactly 1 received a special accommodation? b. What is the probability that at least 1 received a special accommodation? c. What is the probability that at least 2 received a special accommodation? d. What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated? e. Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed \(4.5\) hours. What would you expect the average time allowed the 25 selected students to be?

A trial has just resulted in a hung jury because eight members of the jury were in favor of a guilty verdict and the other four were for acquittal. If the jurors leave the jury room in random order and each of the first four leaving the room is accosted by a reporter in quest of an interview, what is the pmf of \(X=\) the number of jurors favoring acquittal among those interviewed? How many of those favoring acquittal do you expect to be interviewed?

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