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A large cable company reports that \(42 \%\) of its customers subscribe to its Internet service, \(32 \%\) subscribe to its phone service, and \(51 \%\) subscribe to its Internet service or its phone service (or both). a. Use the given probability information to set up a "hypothetical \(1000 "\) table. b. Use the table to find the following: i. the probability that a randomly selected customer subscribes to both the Internet service and the phone service. ii. the probability that a randomly selected customer subscribes to exactly one of the two services.

Short Answer

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a. Assume there are 1000 customers. Based on the given proportions: 420 subscribe to the Internet service, 320 subscribe to the phone service, and 510 subscribe to either the Internet service or the phone service (or both). Then, it can be calculated that 230 customers subscribe to both services, and 280 customers subscribe to exactly one service. b. Therefore, the probability that a randomly selected customer subscribes to both the Internet service and the phone service is 0.23, and the probability that a randomly selected customer subscribes to exactly one of the two services is 0.28.

Step by step solution

01

Create a hypothetical 1000 table

Start by assuming there are 1000 customers (this makes working with percentages easier). Based on the given proportions: \n\n- 420 customers subscribe to the Internet service (42% of 1000), \n\n- 320 customers subscribe to the phone service (32% of 1000), and \n\n- 510 customers subscribe to either the Internet service or the phone service (or both) (51% of 1000).
02

Calculate the number of customers who subscribe to both services

Next, calculate the number of customers that subscribe to both services. This can be obtained by adding the number of customers who subscribe to the Internet and phone service, and then subtracting the number of customers who subscribe to either one. This gives \(420 (Internet service) + 320 (phone service) - 510 (either service) = 230 (both services). \) So, 230 out of 1000 customers subscribe to both services.
03

Calculate the number of customers who subscribe to exactly one service

You can calculate the number of customers who subscribe to exactly one service by subtracting the number of customers who subscribe to both services from the total number of customers who subscribe to either service. This gives \(510 (either service) - 230 (both services) = 280 (exactly one service)\). Therefore, 280 out of 1000 customers subscribe to exactly one service.
04

Convert the customer numbers into probabilities

Finally, you convert these numbers into probabilities. For the probability that a customer subscribes to both services, divide the number of customers who subscribe to both services by the total number of customers, which gives 230/1000 = 0.23. For the probability that a customer subscribes to exactly one service, divide the number of customers who subscribe to exactly one service by the total number of customers, which gives 280/1000 = 0.28.

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Key Concepts

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

Hypothetical 1000 Table
Understanding probability can sometimes be challenging, but a hypothetical 1000 table simplifies it. Imagine that you have a group of 1000 people or objects; when we discuss probabilities in percentages, each percentage point represents 10 members of this hypothetical group.

For example, if a company reports that 42% of its customers use its Internet service, we visualize this as 420 out of 1000 customers in our hypothetical scenario. Similarly, 32% for phone service becomes 320 out of 1000. Setting up such a table makes it straightforward to calculate combined probabilities and visualize overlapping groups.

When given the proportion of customers using either service, we face potential overlaps. The company reports 51%, or 510 customers using one service or both. Using a table, we can easily subtract the sum of individual services from this figure to find how many use both services. This method also proves invaluable in understanding more complex probability scenarios.
Joint Probability
Joint probability is a key concept when we look at how likely two events are to happen at the same time. Formally, it's the probability that two events, A and B, occur together. To calculate it, we take into account the individual probabilities of each event and any overlap between them.

In our example, the joint probability would represent customers who subscribe to both Internet and phone services. The calculation process involves finding the number of customers with both services — here, 230 out of 1000, or 23%. This value is our joint probability, indicating a significant overlap between the two services. It's also a great example to show how businesses can use joint probabilities to understand their customer base and plan bundles or promotions effectively.
Mutually Exclusive Events
While joint probability deals with events that can occur at the same time, mutually exclusive events cannot. Two events are mutually exclusive if the occurrence of one event means the other cannot occur simultaneously. A classic example would be tossing a coin; it cannot land on both heads and tails at the same time.

In our cable company scenario, if there were a group of customers who could only subscribe to either Internet or phone service, but not both, these would be mutually exclusive events. Knowing this is critical when analyzing probabilities, as it affects the calculations. For events that aren't mutually exclusive (like our Internet and phone services example), the total probability can exceed the probabilities of the individual events, reflecting the overlap. Clearly, the concept of mutually exclusive events is central to a proper understanding of probability and is a cornerstone in interpreting data correctly.

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Most popular questions from this chapter

In an article that appears on the website 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 \(C=\) event that the person tested is actually clean a. Using the information in the quote, what are the values of i. \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C)\) iv. \(P(D)\) b. Use the probabilities from Part (a) to construct a "hypothetical 1000 " table. c. What is the value of \(P(T D)\) ? d. Use the table to calculate the probability that a person is clean given that the test result is dirty, \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

A rental car company offers two options when a car is rented. A renter can choose to pre-purchase gas or not and can also choose to rent a GPS device or not. Suppose that the events \(A=\) event that gas is pre-purchased \(B=\) event that a GPS is rented are independent with \(P(A)=0.20\) and \(P(B)=0.15\). a. Construct a "hypothetical 1000 " table with columns corresponding to whether or not gas is pre-purchased and rows corresponding to whether or not a GPS is rented. b. Use the table to find \(P(A \cup B)\). Give a long-run relative frequency interpretation of this probability.

Each time a class meets, the professor selects one student at random to explain the solution to a homework problem. There are 40 students in the class, and no one ever misses class. Luke is one of these students. What is the probability that Luke is selected both of the next two times that the class meets? (Hint: See Example 5.8 )

The same issue of The Chronicle for Higher Education referenced in Exercise 5.17 also reported the following information for degrees awarded to Hispanic students by U.S. colleges in the \(2008-2009\) academic year: A total of 274,515 degrees were awarded to Hispanic students. \- 97,921 of these degrees were Associate degrees. \- 129,526 of these degrees were Bachelor's degrees. \- The remaining degrees were either graduate or professional degrees. What is the probability that a randomly selected Hispanic student who received a degree in \(2008-2009\) a. received an associate degree? b. received a graduate or professional degree? c. did not receive a bachelor's degree?

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 \(\mathrm{Alex}\) 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.

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