/*! 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 29 Of the 60 movies reviewed last y... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.)

Short Answer

Expert verified
Based on the calculation, a. The probability that a movie gets thumbs up from Critic 2 given it has been given thumbs up by Critic 1 is \(\frac{10}{15}\) or \(\frac{2}{3}\). b. The probability that a movie does not get thumbs up from Critic 1 given that it hasn't been given thumbs up by Critic 2 is calculated as \(\frac{30}{40}\) or \(\frac{3}{4}\)

Step by step solution

01

Create a two-way table

Begin by creating a two-by-two table to organize the data. The rows represent Critic 1's ratings (thumbs-up or not) and columns represent Critic 2's ratings (thumbs-up or not). Fill in the table with the number of movies each scenario contains.
02

Calculate the total numbers

Calculate the total number of movies for each row and column. Also, calculate the total number of movies that both critics reviewed.
03

Calculate the probability for part (a)

To answer part (a), given that a movie was rated thumbs-up by Critic 1, what is the probability that it also received this rating from Critic 2, divide the number of movies that received thumbs-up from both critics by the total number of thumbs-up by Critic 1, i.e., P(Critic 2 thumbs-up | Critic 1 thumbs-up) = Both Thumbs-Up / Critic 1 Thumbs up.
04

Calculate the probability for part (b)

To answer part (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, divide the number of movies that did not receive thumbs-up from both critics by the total number of not thumbs-up by Critic 2, i.e., P(Critic 1 not thumbs-up | Critic 2 not thumbs-up) = Both Not-Thumbs-Up / Critic 2 Not Thumbs up

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Theory
Probability theory is the mathematical framework for analyzing random phenomena and quantifying how likely events are to occur. In essence, it revolves around the idea of assigning numbers, probabilities, to events that are scaled usually from 0 to 1, where 0 indicates an impossible event, and 1 signifies a certain one. The field is vast, covering various concepts from simple events to complex phenomena involving multiple factors and conditions.

To apply probability theory to our critics' movie review problem, we use the concept of conditional probability. This term refers to the likelihood of an event occurring given that another event has already happened. For example, when thinking about the probability that Critic 2 gave a thumbs-up after knowing that Critic 1 already did, we're invoking the principles of conditional probability.
Two-way Table
A two-way table, also known as a contingency table or cross-tabulation, is a powerful tool for organizing and analyzing categorical data. This type of table showcases the frequency of different combinations of outcomes for two categorical variables. In our movie review example, one categorical variable is the thumbs-up status by Critic 1 (up or not up), and the other is that of Critic 2 (up or not up).

By creating a two-way table, we can easily visualize and calculate joint, marginal, and conditional probabilities, which is crucial for parts (a) and (b) of the original exercise. The table helps clarify the number of movies that fall into each category of the thumbs-up/thumbs-down relationship between the two critics.
Statistical Independence
Statistical independence is a key concept in probability theory. It denotes a scenario where the occurrence of one event does not affect the probability of another. In simple terms, two events, A and B, are independent if knowing whether A occurred does not change the likelihood of B occurring, and vice versa.

In our context, if the opinions of the two critics were statistically independent, the thumbs-up rating of one critic would not influence the likelihood of the other giving a thumbs-up too. However, the exercise doesn't explicitly state whether the critics' ratings are independent or not. In real-world scenarios, such independence often requires verification through statistical tests or an understanding of the underlying processes.
Joint Probability
Joint probability refers to the probability of two events occurring simultaneously. It’s denoted as P(A and B) or P(A, B), and is easy to work with when using a two-way table. In our movie review example, the joint probability of both critics giving a thumbs-up is derived by dividing the number of movies that both critics liked by the total number of movies reviewed.

For conditional probability calculations, joint probability plays a critical role as well. It is the linking piece between the probabilities of each individual event and the combined outcome of interest. For example, in the case of the critics, knowing the joint probability of both giving a thumbs-up is imperative to answer questions like part (a) of the exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\) and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F ?\) b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F) ?\) A Venn diagram might help.)

Refer to Exercise 6.18. Adding probabilities in the first row of the given table yields \(P(\) midsize \()=.45\), whereas from the first column, \(\mathrm{P}\left(4 \frac{3}{8}\right.\) in. grip) \(=.30\). Is the following true? $$ P\left(\text { midsize } \text { or } 4 \frac{3}{8} \text { in. grip }\right)=.45+.30=.75 $$ Explain.

A transmitter is sending a message using a binary code, namely, a sequence of 0's and 1's. Each transmitted bit \((0\) or 1\()\) must pass through three relays to reach the receiver. At each relay, the probability is \(.20\) that the bit sent on is different from the bit received (a reversal). Assume that the relays operate independently of one another: transmitter \(\rightarrow\) relay \(1 \rightarrow\) relay \(2 \rightarrow\) relay \(3 \rightarrow\) receiver a. If a 1 is sent from the transmitter, what is the probability that a 1 is sent on by all three relays? b. If a 1 is sent from the transmitter, what is the probability that a 1 is received by the receiver? (Hint: The eight experimental outcomes can be displayed on a tree diagram with three generations of branches, one generation for each relay.)

N.Y. Lottery Numbers Come Up 9-1-1 on 9/11" was the headline of an article that appeared in the San Francisco Chronicle (September 13,2002 ). More than 5600 people had selected the sequence \(9-1-1\) on that date, many more than is typical for that sequence. A professor at the University of Buffalo is quoted as saying, "I'm a bit surprised, but I wouldn't characterize it as bizarre. It's randomness. Every number has the same chance of coming up." a. The New York state lottery uses balls numbered \(0-9\) circulating in three separate bins. To select the winning sequence, one ball is chosen at random from each bin. What is the probability that the sequence \(9-1-1\) is the sequence selected on any particular day? (Hint: It may be helpful to think about the chosen sequence as a threedigit number.) b. What approach (classical, relative frequency, or subjective) did you use to obtain the probability in Part (a)? Explain.

Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let \(\mathrm{D}=\) defective and \(\mathrm{N}=\) not defective). The chance experiment terminates as soon as a nondefective battery is obtained. a. Give five possible experimental outcomes. b. What can be said about the number of outcomes in the sample space? c. What outcomes are in the event \(E\), that the number of batteries examined is an even number?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.