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For a monthly subscription fee, a video download site allows people to download and watch up to five movies per month. Based on past download histories, the following table gives the estimated probabilities that a randomly selected subscriber will download 0,1,2,3,4 or 5 movies in a particular month. $$ \begin{array}{|lcccccc|} \hline \text { Number of downloads } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Estimated probability } & 0.03 & 0.45 & 0.25 & 0.10 & 0.10 & 0.07 \\ \hline \end{array} $$ If a subscriber is selected at random, what is the estimated probability that this subscriber downloads a. three or fewer movies? b. at most three movies? c. four or more movies? d. zero or one movie? e. more than one movie?

Short Answer

Expert verified
a) The estimated probability of a randomly selected subscriber downloading three or fewer movies is \(0.83\). b) The estimated probability of a randomly selected subscriber downloading at most three movies is \(0.83\). c) The estimated probability of a randomly selected subscriber downloading four or more movies is \(0.17\). d) The estimated probability of a randomly selected subscriber downloading zero or one movie is \(0.48\). e) The estimated probability of a randomly selected subscriber downloading more than one movie is \(0.52\).

Step by step solution

01

Identify the relevant probabilities

For a randomly selected subscriber downloads three or fewer movies, we are interested in the probabilities for 0, 1, 2, and 3 movies. From the table, these are 0.03, 0.45, 0.25, and 0.10.
02

Calculate the probability of downloading three or fewer movies

To find the probability of downloading three or fewer movies, we will add the probabilities of downloading 0, 1, 2, and 3 movies, which are 0.03, 0.45, 0.25, and 0.10. \(P(\text{three or fewer movies}) = 0.03 + 0.45 + 0.25 + 0.10 = 0.83\) b) At most three movies: This is the same as "three or fewer movies," so the answer is the same as in part a: 0.83. c) Four or more movies:
03

Identify the relevant probabilities

For a randomly selected subscriber downloads four or more movies, we are interested in the probabilities for 4 and 5 movies. From the table, these are 0.10 and 0.07.
04

Calculate the probability of downloading four or more movies

To find the probability of downloading four or more movies, we will add the probabilities of downloading 4 and 5 movies, which are 0.10 and 0.07. \(P(\text{four or more movies}) = 0.10 + 0.07 = 0.17\) d) Zero or one movie:
05

Identify the relevant probabilities

For a randomly selected subscriber downloads zero or one movie, we are interested in the probabilities for 0 and 1 movies. From the table, these are 0.03 and 0.45.
06

Calculate the probability of downloading zero or one movie

To find the probability of downloading zero or one movie, we will add the probabilities of downloading 0 and 1 movies, which are 0.03 and 0.45. \(P(\text{zero or one movie}) = 0.03 + 0.45 = 0.48\) e) More than one movie:
07

Calculate the probability of downloading one or fewer movies

We have already calculated the probability of downloading zero or one movie in part d, which is 0.48.
08

Calculate the probability of downloading more than one movie

To find the probability of downloading more than one movie, we subtract the probability of downloading one or fewer movies from the total probability of 1. \(P(\text{more than one movie}) = 1 - P(\text{one or fewer movies}) = 1 - 0.48 = 0.52\)

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Key Concepts

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

Probability Calculation
Understanding probability calculations is fundamental in statistics and various real-world applications. Probability represents the likelihood of an event happening, and it is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

In the exercise given, we determine the probability of different movie download scenarios for a video service subscriber. To calculate these probabilities, we need to sum the probabilities of the individual events that make their conditions true. For example, the probability of a subscriber downloading three or fewer movies is calculated by adding the individual probabilities for 0, 1, 2, and 3 downloads. The formula is expressed as follows: \[\begin{equation}P(\text{three or fewer movies}) = P(0) + P(1) + P(2) + P(3)\tag{1}\end{equation}\] using the values from the probability table provided.

Effective probability calculation requires careful consideration of what constitutes success for the event under consideration, identifying relevant outcomes, and summing their probabilities if they are mutually exclusive events—which means they cannot happen at the same time.
Statistical Probability
Statistical probability involves using data from previous events, or samples, to estimate the probability of future events. It is a powerful tool that can inform predictions and decision-making processes.

In the context of our exercise, statistical probability is used to estimate the chances of a randomly selected subscriber downloading a certain number of movies based on past download histories. These estimates come from a frequency distribution—a summary of how often each possible outcome occurred in the past.

The estimated probability for each number of downloads, given in the problem, directly reflects the real-world observations. For example, the 0.45 probability for downloading one movie tells us that, in the past, about 45% of subscribers downloaded exactly one movie in a month. This use of historical data is a cornerstone of statistical probability, as it assumes that past behaviors are indicative of future actions, within certain limits.
Probability Distribution
A probability distribution details the likelihood of various outcomes within a set. For discrete random variables like the number of movies downloaded, we use a probability mass function (PMF) to express the probability distribution. Every possible outcome has a probability associated with it, and the sum of all probabilities equals 1.

The exercise presents a type of probability distribution with the probabilities for 0 to 5 movies downloaded. While calculating specific probabilities, understanding distribution helps in grasping why the probabilities in the table must sum to 1. This concept confirms that one of the possible outcomes in the distribution must occur.

Moreover, probability distributions are key when determining the likeliness of a range of outcomes. For instance, calculating the chance of 'zero or one movie' or 'four or more movies' downloaded requires adding the probabilities for those specific ranges. The PMF gives us the probability for each data point, so we can add them to get the probabilities for a range of values, as in:\[\begin{equation}P(\text{four or more movies}) = P(4) + P(5)\tag{2}\end{equation}\] Thus, having a grasp on probability distribution eases the process of calculating cumulative probabilities for given ranges of outcomes.

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Most popular questions from this chapter

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a time line is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is 0.8 . 2\. If Maria completes her part on time, the probability that Alex completes on time is 0.9 , but if Maria is late, the probability that Alex completes on time is only \(0.6 .\) 3\. If Alex completes his part on time, the probability that Juan completes on time is 0.8 , but if Alex is late, the probability that Juan completes on time is only \(0.5 .\) 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(0.9,\) but if Juan is late, the probability that Jacob completes on time is only 0.7 . Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time, and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 4)=0.8 $$ \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 3)=0.7 $$ \(P(\) Seed 2 defeats Seed 3\()=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats \(\operatorname{Seed} 4)=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.

If you were to roll a fair die 1000 times, about how many sixes do you think you would observe? What is the probability of observing a six when a fair die is rolled?

Phoenix is a hub for a large airline. Suppose that on a particular day, 8000 passengers arrived in Phoenix on this airline. Phoenix was the final destination for 1800 of these passengers. The others were all connecting to flights to other cities. On this particular day, several inbound flights were late, and 480 passengers missed their connecting flight. Of these 480 passengers, 75 were delayed overnight and had to spend the night in Phoenix. Consider the chance experiment of choosing a passenger at random from these 8000 passengers. Calculate the following probabilities: a. the probability that the selected passenger had Phoenix as a final destination. b. the probability that the selected passenger did not have Phoenix as a final destination. c. the probability that the selected passenger was connecting and missed the connecting flight. d. the probability that the selected passenger was a connecting passenger and did not miss the connecting flight. e. the probability that the selected passenger either had Phoenix as a final destination or was delayed overnight in Phoenix. f. An independent customer satisfaction survey is planned. Fifty passengers selected at random from the 8000 passengers who arrived in Phoenix on the day described above will be contacted for the survey. The airline knows that the survey results will not be favorable if too many people who were delayed overnight are included. Write a few sentences explaining whether or not you think the airline should be worried, using relevant probabilities to support your answer.

A man who works in a big city owns two cars, one small and one large. Three- quarters of the time he drives the small car to work, and one-quarter of the time he takes the large car. If he takes the small car, he usually has little trouble parking and so is at work on time with probability 0.9. If he takes the large car, he is on time to work with probability 0.6. Given that he was at work on time on a particular morning, what is the probability that he drove the small car?

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