/*! 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 27 Suppose that a six-sided die is ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to be observed as any particular odd-numbered face. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p\), \(P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p .\) Now use a condi- tion on the sum of these probabilities to determine \(p .\) ) b. What is the probability that the number showing is an odd number? at most \(3 ?\) c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots, P\left(O_{6}\right)=6 c .\) What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

Short Answer

Expert verified
a. The 6 probabilities are \( 1/9, 2/9, 1/9, 2/9, 1/9, 2/9 \)\nb. The probability of an odd number is \( 1/3 \), and of a number at most 3 is \( 4/9 \) \nc. The 6 probabilities are \( 1/21, 2/21, 3/21, 4/21, 5/21, 6/21 \). The probability of an odd number is \( 3/7 \), and of a number at most 3 is \( 2/7 \)

Step by step solution

01

Determine the value of \( p \)

The six simple events are \( O_{1} \), \( O_{2} \), \( O_{3} \), \( O_{4} \), \( O_{5} \), \( O_{6} \). Given that \( P(O_{1}) = p \), \( P(O_{2}) = 2p \), \( P(O_{3}) = p \), \( P(O_{4}) = 2p \), \( P(O_{5}) = p \), \( P(O_{6}) = 2p \), we use the fact that the sum of probabilities of all simple events is always equal to 1 to find \( p \). Adding these up, we get \( 9p = 1 \). Solving for \( p \) we find \( p = 1/9 \).
02

Calculate the probability of odd and at most 3

For an odd outcome, we sum the probabilities of \( O_{1} \), \( O_{3} \), and \( O_{5} \), giving us \( p + p + p = 3p \). Substituting the value of \( p \) from Step 1, we get \( 3/9 = 1/3 \). For an outcome of at most 3, we sum the probabilities of \( O_{1} \), \( O_{2} \), and \( O_{3} \), giving us \( p + 2p + p = 4p \). Substituting the value of \( p \) from Step 1, we get \( 4/9 \).
03

Determine the value of \( c \)

Here, the probability of each outcome is proportional to the number on the face, denoted as \( c, 2c, 3c, 4c, 5c, 6c \). Adding these up and equating to 1, we get \( 21c = 1 \). Solving for \( c \), we get \( c = 1/21 \).
04

Calculate new probabilities

For an odd outcome with the new dice, we sum the probabilities of \( O_{1} \), \( O_{3} \), and \( O_{5} \), giving us \( c + 3c + 5c = 9c = 9/21 = 3/7 \). For an outcome of at most 3 with the new dice, we sum the probabilities of \( O_{1} \), \( O_{2} \), and \( O_{3} \), giving us \( c + 2c + 3c = 6c = 6/21 = 2/7 \).

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Key Concepts

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

Loaded Die
A loaded die refers to a die that is not fair, meaning the outcomes are not equally likely. In our case, the die is loaded such that even-numbered faces appear more often than odd-numbered ones.
This loading affects probability calculations and can be intentional for games or simulations. By altering the likelihood of outcomes, you can model scenarios closer to what you’re studying or ensure more frequent specific outcomes.
Understanding a loaded die involves recognizing that the die does not adhere to the usual uniform probability distribution. Every face of a fair die has an equal chance (i.e., 1/6) of appearing. Here, balance is disrupted, with even faces being twice as likely as odd ones. This concept helps us calculate probabilities differently, as shown in the exercise by denoting probabilities like \( P(O_1) = p \) and \( P(O_2) = 2p \). Understanding this setup illustrates how probabilities are adjusted to account for the 'loaded' characteristic of the die.
Simple Events
Simple events are fundamental components in probability, representing the outcome of a random experiment. Here, each side of the die throw is a simple event.
When rolling a six-sided die, the simple events are the occurrence of one of the six numbers: 1, 2, 3, 4, 5, or 6. Each outcome you've denoted as \( O_1, O_2, O_3, O_4, O_5, O_6 \).
These simple events are critical in probability because they form the building blocks to understand more complex questions. For instance, calculating probabilities for a combination of these simple events, such as finding the chance of rolling an odd number or a number at most 3. Simple events make it possible to decompose complex probability questions into manageable parts.
  • Each die face is a simple event.
  • Probabilities assigned to these events depend on whether the die is fair or loaded.
  • Used to sum probabilities in complex calculations.
Proportional Probability
Proportional probability refers to assigning probabilities based on the proportion of some quantitative factor. In this exercise, the probability of each outcome corresponds to the face value of the die.
This means the number '1' has a probability \( c \), the number '2' has \( 2c \), and so on, making the outcome of rolling a '6' more likely than rolling a '1'. Such proportional approaches adjust probability calculations to fit specific criteria or datasets.
This principle is crucial when a non-uniform event distribution is needed, often used in weighted models. The condition \( 21c = 1 \) was used to find the constant \( c = 1/21 \), ensuring that probabilities for all dice outcomes sum up to 1. It highlights how increasing or decreasing proportional factors can alter total probabilities, effectively customizing your model.
Odds and Evens
In probability, distinguishing between odd and even outcomes can simplify calculations for certain questions, as they split possibilities in a binary manner.
In this exercise, odd numbers are 1, 3, and 5, and even numbers are 2, 4, and 6. By grouping numbers into odds and evens, you can swiftly calculate their combined probabilities, which is essential when working with loaded dice scenarios.
When focusing on outcomes like odd numbers, the probability combines all odd-numbered events: \( P(O_1) + P(O_3) + P(O_5) \). Simplifyingly, it can provide easier answers to specific questions without dealing with all distinct outcomes individually. This strategy helps manage complexity in probability exercises, ensuring results are easily obtainable and understandable while maintaining accuracy.

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Most popular questions from this chapter

The Associated Press (San Luis Obispo TelegramTribune, August 23,1995 ) reported on the results of mass screening of schoolchildren for tuberculosis (TB). For Santa Clara County, California, the proportion of all tested kindergartners who were found to have TB was .0006. The corresponding proportion for recent immigrants (thought to be a high-risk group) was .0075. Suppose that a Santa Clara County kindergartner is selected at random. Are the events selected student is a recent immigrant and selected student has \(T B\) independent or dependent events? Justify your answer using the given information.

Of the 60 movies reviewed last year by two critics on their joint television show, Critic 1 gave a "thumbs-up" rating to 15 , Critic 2 gave this rating to 20 , and 10 of the movies were rated thumbs-up by both critics. Suppose that 1 of these 60 movies is randomly selected. a. Given that the movie was rated thumbs-up by Critic 1 , what is the probability that it also received this rating from Critic \(2 ?\) b. If the movie did not receive a thumbs-up rating from Critic 2, what is the probability that it also did not receive a thumbs up rating from Critic \(1 ?\) (Hint: Construct a table with two rows for the first critic [for "up" and "not up"] and two columns for the second critic: then enter the relevant probabilities.)

In an article that appears on the web site of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 \- are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty, \(T C=\) event that the test result is clean, \(D=\) event that the person tested is actually dirty, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C) \quad\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

There are five faculty members in a certain academic department. These individuals have \(3,6,7,10\), and 14 years of teaching experience. Two of these individuals are randomly selected to serve on a personnel review committee. What is the probability that the chosen representatives have a total of at least 15 years of teaching experience? (Hint: Consider all possible committees.)

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7, P(F)=.6\), and \(P(E \cap F)=.54\). a. Calculate \(P(E \mid F)\) the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

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