/*! 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 35 Consider the population of voter... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the population of voters described in Example \(3.6 .\) Suppose that there are \(N=5000\) voters in the population, \(40 \%\) of whom favor Jones. Identify the event favors Jones as a success \(S\). It is evident that the probability of \(S\) on trial 1 is .40 . Consider the event \(B\) that \(S\) occurs on the second trial. Then \(B\) can occur two ways: The first two trials are both successes or the first trial is a failure and the second is a success. Show that \(P(B)=.4 .\) What is \(P(B | \text { the first trial is } S\) )? Does this conditional probability differ markedly from \(P(B) ?\)

Short Answer

Expert verified
\(P(B | \text{first trial is } S) = 0.40\), same as \(P(B)\).

Step by step solution

01

Define Events and Probabilities

Identify the event 'favors Jones' as a success \(S\). The probability of success on any trial is \(P(S) = 0.40\). The complementary event, failure, is denoted \(F\) where \(P(F) = 0.60\) since \(P(F) = 1 - P(S)\).
02

Identify Ways Event B Occurs

There are two scenarios for event \(B\) (a success on the second trial): (i) both trials are successes (\(S, S\)), or (ii) the first trial is a failure and the second trial a success (\(F, S\)).
03

Calculate Probability for Both Trials Success (S, S)

The probability that both trials are successes is \(P(S) \times P(S) = 0.40 \times 0.40 = 0.16\).
04

Calculate Probability for First Fail and Second Success (F, S)

The probability for the first trial being a failure and the second a success is \(P(F) \times P(S) = 0.60 \times 0.40 = 0.24\).
05

Compute Total Probability for Event B

Add the probabilities from the two scenarios to find \(P(B)\): \(P(B) = P(S, S) + P(F, S) = 0.16 + 0.24 = 0.40\).
06

Calculate Conditional Probability P(B | S)

Given the first trial is a success (\(S\)), event \(B\) happens if the second trial is also a success. Thus, \(P(B | \text{first trial is } S) = P(S) = 0.40\).
07

Compare P(B) and P(B | S)

Both \(P(B)\) and \(P(B | \text{first trial is } S)\) are 0.40. The conditional probability does not differ from the original probability of event \(B\).

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Key Concepts

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

Probability Calculations
Understanding probability calculations is crucial in the context of making predictions about the outcome of an event. In this exercise, we are dealing with basic probability, where we have two possible outcomes: a success or a failure. The probability, denoted by \( P(S) \), of an event being a success (favoring Jones) is 0.40.
Probability calculations are dependent on known values and assumptions. We assume that each trial is independent and that the probability of success remains constant at 0.40 throughout the trials. We use this probability to calculate the likelihood of various outcomes. For instance, the probability of both trials being successes is calculated by multiplying \( P(S) \) for two trials, resulting in 0.16.
This approach offers a structured method for determining outcomes when dealing with probability, serving as a foundation for more complex calculations seen in joint probabilities and conditional probabilities.
Law of Total Probability
The Law of Total Probability is a fundamental rule that provides a way to calculate the probability of an event based on the probabilities of other disjoint events. In our exercise, we determine \( P(B) \), the probability of a success on the second trial, using this law.
Event \( B \), a success on trial two, can occur through two distinct scenarios: both trials succeeding or only the second trial succeeding after the first one fails. We calculate each scenario's probability first, adding them up to get the total probability of \( B \).
This law helps break down complex probability problems into simpler parts. Calculating the scenarios separately allows for an easier and clearer approach to finding the total probability of an event occurring, thus ensuring accurate results in various probability calculations.
Binomial Experiments
Binomial experiments are statistical experiments that have a fixed number of independent trials, each with the same probability of a particular outcome, such as success or failure. This exercise is a great example of how binomial experiments work in practice.
Each voter's decision is treated as a separate trial where success means the voter favors Jones. Given the probability of success \( P(S) = 0.40 \), we navigate through the process of determining the probability of a successful outcome in these trials.
  • A fixed number of trials (such as several voters making decisions).
  • Two possible outcomes per trial (success or failure).
  • Consistent probability of success for each trial.
Understanding these elements helps in modeling events like elections, roll of dice, or quality testing in manufacturing. Calculating probabilities using a binomial setup provides students with essential tools for interpreting various statistical data.

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Most popular questions from this chapter

Let \(Y\) denote a geometric random variable with probability of success \(p\) a. Show that for a positive integer \(a\) $$P(Y>a)=q^{a}$$, b. Show that for positive integers \(a\) and \(b\) ,$$P(Y>a+b | Y>a)=q^{b}=P(Y>b)$$ This result implies that, for example, \(P(Y>7 | Y>2)=P(Y>5) .\) Why do you think this property is called the memoryless property of the geometric distribution? c. In the development of the distribution of the geometric random variable, we assumed that the experiment consisted of conducting identical and independent trials until the first success was observed. In light of these assumptions, why is the result in part (b) "obvious"?

A missile protection system consists of \(n\) radar sets operating independently, each with a probability of .9 of detecting a missile entering a zone that is covered by all of the units. a. If \(n=5\) and a missile enters the zone, what is the probability that exactly four sets detect the missile? At least one set? b. How large must \(n\) be if we require that the probability of detecting a missile that enters the zone be \(.999 ?\)

Construct probability histograms for the binomial probability distributions for \(n=5\) \(p=.1, .5,\) and. \(9 .\) (Table \(1,\) Appendix 3 , will reduce the amount of calculation.) Notice the symmetry for \(p=.5\) and the direction of skewness for \(p=.\). 1and. \(.9 .\)

Sampling for defectives from large lots of manufactured product yields a number of defectives, \(Y\), that follows a binomial probability distribution. A sampling plan consists of specifying the number of items \(n\) to be included in a sample and an acceptance number a. The lot is accepted if \(Y \leq a\) and rejected if \(Y>a\). Let \(p\) denote the proportion of defectives in the lot. For \(n=5\) and \(a=0\), calculate the probability of lot acceptance if a. \(p=0\) b. \(p=.1\) C. \(p=.3\) d. \(p=.5\) e. \(p=1.0\) A graph showing the probability of lot acceptance as a function of lot fraction defective is called the operating characteristic curve for the sample plan. Construct the operating characteristic curve for the plan \(n=5, a=0 .\) Notice that a sampling plan is an example of statistical inference. Accepting or rejecting a lot based on information contained in the sample is equivalent to concluding that the lot is either good or bad. "Good" implies that a low fraction is defective and that the lot is therefore suitable for shipment.

Let \(Y\) denote a random variable that has a Poisson distribution with mean \(\lambda=2 .\) Find a. \(P(Y=4)\) b. \(P(Y \geq 4)\) c. \(P(Y < 4)\) d. \(P(Y \geq 4 | Y \geq 2)\)

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