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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

5.57 There are two traffic lights on Shelly's route from home to work. Let \(E\) denote the event that Shelly must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=0.4, P(F)=0.3\) and \(P(E \cap F)=0.15 .\) In Exercise \(5.25,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that Shelly must stop for at least one light (the probability of the event \(E \cup F\) ). b. The probability that Shelly does not have to stop at either light. c. The probability that Shelly must stop at exactly one of the two lights. d. The probability that Shelly must stop only at the first light.

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. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.64\) and \(P(F)=0.17\) i. What is the value of \(P(E \cap F)\) ? ii. What is the value of \(P(E \cup F)\) ? b. Suppose that \(A\) and \(B\) are events with \(P(A)=0.3, P(B)=0.5\), and \(P(A \cap B)=0.15 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell? c. Suppose that \(A\) and \(B\) are events with \(P(A)=0.65\) and \(P(B)=0.57 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

Lyme disease is transmitted by infected ticks. Several tests are available for people with symptoms of Lyme disease. One of these tests is the EIA/IFA test. The paper "Lyme Disease Testing by Large Commercial Laboratories in the United States" (Clinical Infectious Disease [2014]: \(676-681\) ) found that \(11.4 \%\) of those tested actually had Lyme disease. Consider the following events: \(+\) represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities are based on percentages given in the paper: $$ \begin{array}{r} P(L)=0.114 \\ P\left(L^{C}\right)=0.886 \end{array} $$ $$ \begin{array}{c} P(+\mid L)=0.933 \\ P(-\mid L)=0.067 \\ P\left(+\mid L^{C}\right)=0.039 \\ P\left(-\mid L^{C}\right)=0.961 \end{array} $$ a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a hypothetical 1000 table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use information from the table constructed in Part (b) to calculate this probability.

Consider a chance experiment that consists of selecting a customer at random from all people who purchased a car at a large car dealership during 2016 . a. In the context of this chance experiment, give an example of two events that would be mutually exclusive. b. In the context of this chance experiment, give an example of two events that would not be mutually exclusive.
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