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Chapter 5: Problem 26

A large cable company reports that \(80 \%\) of its customers subscribe to its cable TV service, \(42 \%\) subscribe to its Internet service, and \(97 \%\) subscribe to at least one of these two services. a. Use the given probability information to set up a hypothetical 1000 table. b. Use the table from Part (a) to find the following probabilities: i. the probability that a randomly selected customer subscribes to both cable TV and Internet service. ii. the probability that a randomly selected customer subscribes to exactly one of these services.

Short Answer

Expert verified
In summary, for a large cable company with 1000 customers, given that 80% subscribe to cable TV, 42% subscribe to internet and 97% subscribe to at least one: - The probability a randomly selected customer subscribes to both cable TV and Internet service is \(0.25\) or \(25\%\). - The probability a randomly selected customer subscribes to exactly one of these services is \(0.72\) or \(72\%\).

Step by step solution

01

Determine number of customers

We are given that out of 1000 customers: - 80% of customers subscribe to cable TV service - 42% of customers subscribe to internet service - 97% of the customers subscribe to at least one of these services Let's represent the values as: - A: Number of customers who subscribes to cable TV service - B: Number of customers who subscribes to Internet service - T: Total number of customers We have, T = 1000. We need to calculate the customers who subscribe to both services, so we can use this formula: \[A \cap B = A + B - (A \cup B)\] Where: - \(A \cap B\) is the number of customers who subscribe to both services. - \(A \cup B\) is the number of customers who subscribe to at least one service. Now, let's plug in the values: \(A = 0.8 \times T = 0.8 \times 1000 = 800\) \(B = 0.42 \times T = 0.42 \times 1000 = 420\) \(A \cup B= 0.97 \times T = 0.97 \times 1000 = 970\) Plugging these into the formula: \(A \cap B = 800 + 420 - 970 = 250\) Now we have the number of customers who subscribe to both services.
02

Create the 1000 table

Now we'll create the 1000 table based on the information we calculated in Step 1: | | Cable TV Service | No Cable TV Service | Total | |---------------------|------------------|---------------------|-------| | Internet Service | 250 | 170 | 420 | | No Internet Service | 550 | 30 | 580 | | Total | 800 | 200 | 1000 |
03

Calculate Probability for Part i

Now that we have our 1000 table, we can answer Part i, which asks for the probability that a randomly selected customer subscribes to both cable TV and Internet service. \[P(A \cap B) = \frac{A \cap B }{T} = \frac{250}{1000} = 0.25\] Thus, the probability that a randomly selected customer subscribes to both cable TV and Internet service is 0.25 or 25%.
04

Calculate Probability for Part ii

Part ii asks for the probability that a randomly selected customer subscribes to exactly one of these services. From the table, there are 170 customers who subscribe to Internet service only and 550 customers who subscribe to cable TV service only. We can calculate the probability as follows: \[P(Exactly One Service) = \frac{170 + 550}{1000} = \frac{720}{1000} = 0.72\] Thus, the probability that a randomly selected customer subscribes to exactly one of these services is 0.72 or 72%.

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Key Concepts

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

Joint Probability
Joint probability is a key concept in understanding how often two events occur together. For instance, in the exercise where we are looking to find the probability that a customer subscribes to both cable TV and Internet service, we are concerned with joint probability.

Mathematically, joint probability is expressed as the likelihood of two intersecting events occurring at the same time, denoted by the symbol \(P(A\cap B)\). It is calculated as the number of outcomes where both events occur divided by the total number of possible outcomes. In the exercise, this translates to 250 out of 1000 customers subscribed to both services, leading to a joint probability of 25%. When understanding joint probability, it is crucial to recognize that it requires both events to happen simultaneously.
Complement Rule
The complement rule is incredibly useful to find the probability of the event not happening. Essentially, the complement rule states that the sum of the probability of an event and the probability of its complement (the event not happening) is equal to 1.

For instance, if we want to find the probability that a customer does not subscribe to either cable TV or Internet service, we would use the complement of subscribing to at least one service. Therefore, if 97% subscribe to at least one service, the complement, those subscribing to neither, would be \(1 - 0.97 = 0.03\) or 3%. This rule is perfect for flipping the perspective from occurrence to non-occurrence and is expressed as \(P(A^c) = 1 - P(A)\).
Conditional Probability
Conditional probability refers to the likelihood of an event occurring, given that another event has already occurred. This parameter is useful when the outcome of one event influences the outcome of another. It's denoted by \(P(A|B)\), and it's distinct because it takes into consideration the effect of event B on event A.

In the context of our cable company example, if we wanted to know the probability that a customer subscribes to cable TV service given that they already subscribe to Internet service, we would be dealing with conditional probability. It requires careful analysis of the shared outcomes relative to the total outcomes of the given condition.
Independent Events
Independent events are those whose occurrence does not affect each other. In probability terms, two events A and B are independent if the occurrence of A does not alter the probability of B occurring, and vice versa.

Mathematically, this is represented as \(P(A \cap B) = P(A) \times P(B)\), when both events are independent. For our exercise, the probability of the cable company customers subscribing to both cable TV and Internet service is not specifically stated to be independent but is calculated through the given probabilities. In real-world scenarios, independent events simplify probability calculations but it's imperative to verify independence before applying such simplifications.

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Most popular questions from this chapter

The report "Improving Undergraduate Learning" (Social Science Research Council, 2011) summarizes data from a survey of several thousand college students. These students were thought to be representative of the population of all college students in the United States. When asked about an upcoming semester, \(68 \%\) said they would be taking a class that is reading-intensive (requires more than 40 pages of reading per week). Only \(50 \%\) said they would be taking a class that is writing-intensive (requires more than 20 pages of writing over the course of the semester). The percentage who said that they would be taking both a reading-intensive course and a writing-intensive course was \(42 \%\). a. Use the given information to set up a hypothetical 1000 table. b. Use the table to find the following probabilities: i. the probability that a randomly selected student would be taking at least one reading-intensive or writing-intensive course. ii. the probability that a randomly selected student would be taking a reading-intensive course or a writingintensive course, but not both. iii. the probability that a randomly selected student would be taking neither a reading-intensive nor a writing-intensive course.

Consider the following two lottery-type games: Game 1 : You pick one number between 1 and \(50 .\) After you have made your choice, a number between 1 and 50 is selected at random. If the selected number matches the number you picked, you win. Game 2 : You pick two numbers between 1 and \(10 .\) After you have made your choices, two different numbers between 1 and 10 are selected at random. If the selected numbers match the two you picked, you win. a. The cost to play either game is \(\$ 1,\) and if you win you will be paid \(\$ 20 .\) If you can only play one of these games, which game would you pick and why? Use relevant probabilities to justify your choice. b. For either of these games, if you plan to play the game 100 times, would you expect to win money or lose money overall? Explain.

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.

An airline reports that for a particular flight operating daily between Phoenix and Atlanta, the probability of an on-time arrival is 0.86. Give a relative frequency interpretation of this probability.

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.
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