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Let \(r\) be a binomial random variable representing the number of successes out of \(n\) trials. (a) Explain why the sample space for \(r\) consists of the set \(\\{0,1,2, \ldots, n\\}\) and why the sum of the probabilities of all the entries in the entire sample space must be 1 . (b) Explain why \(P(r \geq 1)=1-P(0)\). (c) Explain why \(P(r \geq 2)=1-P(0)-P(1)\). (d) Explain why \(P(r \geq m)=1-P(0)-P(1)-\ldots-P(m-1)\) for \(1 \leq m \leq n\).

Short Answer

Expert verified
The sample space consists of \(\{0,1,2,\ldots,n\}\) because those are possible successes. \(P(r \geq 1) = 1 - P(0)\) and \(P(r \geq 2) = 1 - P(0) - P(1)\) from complementary rules. Generalizing gives \(P(r \geq m) = 1 - P(0) - P(1) - \ldots - P(m-1)\).

Step by step solution

01

Understanding the Sample Space

A binomial random variable, denoted as \(r\), represents the number of successes in \(n\) independent trials, each with a success probability \(p\). Therefore, the possible number of successes \(r\) can be any integer from 0 to \(n\), inclusive. This is why the sample space of \(r\) is \(\{0, 1, 2, \ldots, n\}\). Since these are all possible outcomes, the sum of the probabilities of all these outcomes must equal 1, because the total probability of all possible outcomes of any experiment must add up to 1.
02

Deriving Probability for at least One Success

The event \(r \geq 1\) means there is at least one success. The complementary event to having at least one success is having zero successes, \(P(0)\). Hence, by the complementary rule of probability, \(P(r \geq 1) = 1 - P(0)\).
03

Deriving Probability for Two or More Successes

The event \(r \geq 2\) indicates having at least two successes. The complementary events are having zero or one success, corresponding to probabilities \(P(0)\) and \(P(1)\), respectively. Therefore, \(P(r \geq 2) = 1 - P(0) - P(1)\) applies, as it sums the probabilities of the complementary events.
04

Generalizing to at least `m` Successes

For the event \(r \geq m\), we want the probability of achieving at least \(m\) successes. The complementary events are having fewer than \(m\) successes, which are represented by \(P(0), P(1), \ldots, P(m-1)\). Therefore, the probability is \(P(r \geq m) = 1 - P(0) - P(1) - \ldots - P(m-1)\) for \(1 \leq m \leq n\), which subtracts the sum of the probabilities of all outcomes with fewer than \(m\) successes.

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Key Concepts

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

Sample Space
A sample space is a fundamental concept in probability theory. It includes all the possible outcomes of an experiment. When dealing with a binomial random variable, denoted as \( r \), the experiment is usually about the number of successes in \( n \) independent trials. Each trial has a probability \( p \) of success.

For instance, if you were to flip a coin three times, the number of heads (successes) can be 0, 1, 2, or 3. Thus, the sample space for the number of successes in our simple coin-flip scenario would be \( \{0, 1, 2, 3\} \). More generally, for any \( n \) trials, the sample space would be \( \{0, 1, 2, \ldots, n\} \).

The key idea is that the sum of all probabilities of these outcomes must add up to 1. This is because one of the outcomes must occur when you conduct the experiment. In our example, the probability of getting zero heads plus the probability of one head plus the probability of two heads plus the probability of three heads equals 1, ensuring all possibilities are accounted for.
Binomial Random Variable
A binomial random variable, denoted as \( r \), captures the idea of counting how many times a specific event occurs (successes) during a set number of trials. Each separate trial is independent and has two possible outcomes: success or failure. Think of it as flipping a fair coin \( n \) times and counting the number of heads, where each head is considered a success.

The defining characteristics of a binomial random variable include:
  • Fixed number of trials: \( n \), which indicates how many attempts there are.
  • Only two possible outcomes per trial: success (typically labeled as 1) or failure (labeled as 0).
  • Constant probability of success: \( p \), across all trials.
  • Independence of trials: The result of any given trial does not affect others.
Understanding these characteristics helps when calculating probabilities of different numbers of successes through a structured and neat approach, like the binomial formula.
Probability
Probability is essentially a measure of how likely an event is to occur. In the realm of binomial distributions, calculating probabilities involves determining the likelihood of achieving a certain number of successes in a series of independent trials.

Generally, for a binomial distribution, the probability of exactly \( k \) successes in \( n \) trials is given by:\[P(r = k) = \binom{n}{k} p^k (1-p)^{n-k}\]

In this formula:
  • \( \binom{n}{k} \) is the combination of \( n \) things taken \( k \) at a time, calculating the number of ways to achieve \( k \) successes.
  • \( p^k \) represents the probability of \( k \) successes with \( p \) being the probability of a single success.
  • \( (1-p)^{n-k} \) covers the probability of \( n-k \) failures, knowing each trial can only result in a success or failure.
This formula is fundamental to problems involving binomial random variables, enabling you to calculate specific probabilities given a scenario.
Complementary Events
Complementary events are pairs of events, where one event occurs if and only if the other does not. A significant property of complementary events is that their probabilities add up to 1. Therefore, knowing the probability of one event allows us to find the probability of its complement easily.

In binomial distribution, such complementary concepts are handy for calculating cumulative probabilities. For instance:
  • To find \( P(r \geq 1) \), which is the probability of having at least one success, it's often easier to calculate \( 1 - P(0) \). This is because \( P(r \geq 1) \) and \( P(0) \) are complementary events.
  • Similarly, \( P(r \geq 2) \) signifies at least two successes and can be derived by: \( 1 - P(0) - P(1) \). Here, \( P(0) \) and \( P(1) \) are considered as complementing \( P(r \geq 2) \).
These principles make it straightforward to tackle probability questions involving cumulative scenarios in binomial distributions by using the properties of complementary events, simplifying computations and enhancing understanding.

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Most popular questions from this chapter

In an experiment, there are \(n\) independent trials. For each trial, there are three outcomes, \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). For each trial, the probability of outcome \(\mathrm{A}\) is \(0.40\); the probability of outcome \(\mathrm{B}\) is \(0.50\); and the probability of outcome \(\mathrm{C}\) is \(0.10\). Suppose there are 10 trials. (a) Can we use the binomial experiment model to determine the probability of four outcomes of type A, five of type \(\mathrm{B}\), and one of type C? Explain. (b) Can we use the binomial experiment model to determine the probability of four outcomes of type \(\mathrm{A}\) and six outcomes that are not of type A? Explain. What is the probability of success on each trial?

Have you ever tried to get out of jury duty? About \(25 \%\) of those called will find an excuse (work, poor health, travel out of town, etc.) to avoid jury duty (Source: Bernice Kanner, Are You Normal?, St. Martin's Press, New York). If 12 people are called for jury duty, (a) what is the probability that all 12 will be available to serve on the jury? (b) what is the probability that 6 or more will not be available to serve on the jury? (c) Find the expected number of those available to serve on the jury. What is the standard deviation? (d) Quota Problem How many people \(n\) must the jury commissioner contact to be \(95.9 \%\) sure of finding at least 12 people who are available to serve?

An archaeological excavation at Burnt Mesa Pueblo showed that about \(10 \%\) of the flaked stone objects were finished arrow points (Source: Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by Kohler, Washington State University). How many flaked stone objects need to be found to be \(90 \%\) sure that at least one is a finished arrow point? (Hint: Use a calculator and note that \(P(r \geq 1) \geq 0.90\) is equivalent to \(1-P(0) \geq 0.90\), or \(P(0) \leq 0.10 .)\)

The Honolulu Advertiser stated that in Honolulu there was an average of 661 burglaries per 100,000 households in a given year. In the Kohola Drive neighborhood there are 316 homes. Let \(r=\) number of these homes that will be burglarized in a year. (a) Explain why the Poisson approximation to the binomial would be a good choice for the random variable \(r .\) What is \(n\) ? What is \(p ?\) What is \(\lambda\) to the nearest tenth? (b) What is the probability that there will be no burglaries this year in the Kohola Drive neighborhood? (c) What is the probability that there will be no more than one burglary in the Kohola Drive neighborhood? (d) What is the probability that there will be two or more burglaries in the Kohola Drive neighborhood?

Susan is a sales representative who has a history of making a successful sale from about \(80 \%\) of her sales contacts. If she makes 12 successful sales this week, Susan will get a bonus. Let \(n\) be a random variable representing the number of contacts needed for Susan to get the 12 th sale. (a) Explain why a negative binomial distribution is appropriate for the random variable \(n .\) Write out the formula for \(P(n)\) in the context of this application. Hint: See Problem 26 . (b) Compute \(P(n=12), P(n=13)\), and \(P(n=14)\). (c) What is the probability that Susan will need from 12 to 14 contacts to get the bonus? (d) What is the probability that Susan will need more than 14 contacts to get the bonus? (e) What are the expected value \(\mu\) and standard deviation \(\sigma\) of the random variable \(n\) ? Interpret these values in the context of this application.

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