/*! 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 54 Suppose that \(Y\) is a binomial... [FREE SOLUTION] | 91Ó°ÊÓ

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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"?

Short Answer

Expert verified
Y* is a binomial variable with probabilities flipped, showing symmetry in failures and successes.

Step by step solution

01

Understanding the Setup

Consider the random variable \(Y\), which follows a binomial distribution with parameters \(n\) (number of trials) and \(p\) (probability of success in each trial). We define a new variable \(Y^*\) as \(n - Y\), which represents the number of failures in these \(n\) trials.
02

Explaining Part (a)

Part (a) requires us to show that \(P(Y^* = y^*) = P(n - Y = y^*) = P(Y = n - y^*)\). This is directly derived from the definition \(Y^* = n - Y\). Thus, if \(Y^* = y^*\), then \(Y = n - y^*\). The probability statement translates accordingly from the event concerning \(Y^*\) to the event concerning \(Y\).
03

Using the Binomial Probability Formula

For a binomial random variable \(Y\) with parameters \(n\) and \(p\), the probability of having exactly \(y\) successes is given by \(P(Y = y) = \binom{n}{y} p^y (1-p)^{n-y}\).
04

Demonstrating Part (b)

Using the result from part (a), \(P(Y^*=y^*) = P(Y = n-y^*)\). Substitute into the binomial probability formula: \[ P(Y = n-y^*) = \binom{n}{n-y^*} p^{n-y^*} (1-p)^{y^*} \]. Note that \(\binom{n}{n-y^*} = \binom{n}{y^*}\) due to symmetry in combinations. Then, rearrange: \[ P(Y^* = y^*) = \binom{n}{y^*} q^{y^*} p^{n-y^*} \] where \(q = 1-p\).
05

Conclusion for Part (c)

Part (c) asks why \(Y^*\) having a binomial distribution with "success" probability \(p^* = q = 1-p\) is obvious. Since \(Y^*\) counts the number of failures (each with probability \(q = 1 - p\)), it is itself a binomial with \(n\) trials and failure (or inverse success) probability. This aligns with the definition and symmetry of binomial trials.

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Key Concepts

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

Probability
In the context of binomial distribution, probability refers to the likelihood of a specific outcome when an event occurs. A binomial experiment involves a fixed number of trials, often denoted by \(n\), where each trial has two possible outcomes: success or failure. The probability of success in each trial is indicated by \(p\), and the probability of failure, symbolically represented as \(q\), is \(1-p\). This setup forms the foundational structure for evaluating the likelihood of certain results within the framework, such as the probability of having exactly \(k\) successes out of \(n\) trials. To calculate the probability of a given outcome in a binomial distribution, we use the Binomial Probability Formula: \[ P(Y = y) = \binom{n}{y} p^y (1-p)^{n-y} \] Here, \(\binom{n}{y}\) is the binomial coefficient, representing the number of ways to choose \(y\) successes (combinations) from \(n\) trials. The formula effectively combines combinatorial aspects with probability, providing a precise way to ascertain the likelihood of various distributive outcomes. Understanding probability in binomial contexts involves recognizing that each trial is independent, meaning the outcome of one does not affect the others. This principle is intrinsic to calculating the probability for any number of successes or failures.
Success and Failure
In a binomial distribution framework, each trial yields either a success or a failure. The values \(p\) and \(q\) (where \(q = 1-p\)) define the probability of these outcomes in every trial. Success and failure are two complementary aspects – together, they account for all possible outcomes of each trial. A success is what we're interested in counting during the trials. However, due to the nature of binomial distribution, it’s just as crucial to consider the failures, especially because determining the number of failures provides a different perspective. For instance, if you know the number of failures \(Y^*\) (considering \(Y = \text{successes} \)) in \(n\) trials, the number of successes is simply \(n - Y^*\). When we reverse the perspective and focus on failures by considering \(Y^* = n - Y\), the concept of success shifts to counting those failures as our primary interest. Here, the probability \(p^* = q\) becomes the success probability, even though it originally was the probability for a failure in the standard sense of the word. Thus, the binomial trials can be reinterpreted from the angle of observing non-success outcomes, firmly linking success and failure as interchangeable focal points in a binomial experiment.
Combinatorial Analysis
Combinatorial analysis is a crucial aspect of the binomial distribution, as it helps quantify how outcomes can occur. At its core, it involves calculating the number of ways to choose a certain number of successes (or failures) from a total number of trials. This is done using the concept of combinations, specified by the binomial coefficient \(\binom{n}{y}\), which reads "n choose y." Calculating \(\binom{n}{y}\) involves determining how to select \(y\) elements (e.g., successful trials) from \(n\) without regard to order, expressed mathematically as: \[ \binom{n}{y} = \frac{n!}{y!(n-y)!} \] This formula uses factorial notation, where \(!\) denotes the product of all positive integers up to a number, like \(n! = n \times (n-1) \times \ldots \times 1\). Combinatorial analysis enables us to see the symmetry within a binomial distribution. For example, \(\binom{n}{y} = \binom{n}{n-y}\), highlighting that choosing \(y\) successes is equivalently the same as choosing \(n-y\) failures. This inherent symmetry plays a critical role, especially when illustrating identities in binomial probabilities, such as when calculating the probability for \(Y^*\) in terms of \(Y\). It forms the backbone of understanding and manipulating the components of probability tied to the outcomes of binomial trials.

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Most popular questions from this chapter

In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been properly drilled in the boxes and in ten minutes if the holes must be re-drilled. Twenty gearboxes are in stock, 2 with improperly drilled holes. Five gearboxes must be selected from the 20 that are available for installation in the next five robots. a. Find the probability that all 5 gearboxes will fit properly. b. Find the mean, variance, and standard deviation of the time it takes to install these 5 gearboxes.

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