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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

Suppose you want to estimate the probability that a randomly selected customer at a particular grocery store will pay by credit card. Over the past 3 months, 80,500 purchases were made, and 37,100 of them were paid for by credit card. What is the estimated probability that a randomly selected customer will pay by credit card?

A college job placement center has requests from five students for employment interviews. Three of these students are math majors, and the other two students are statistics majors. Unfortunately, the interviewer has time to talk to only two of the students. These two will be randomly selected from among the five. a. What is the sample space for the chance experiment of selecting two students at random? (Hint: You can think of the students as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E}\). One possible selection of two students is \(\mathrm{A}\) and \(\mathrm{B}\). There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that both selected students are statistics majors? d. What is the probability that both students are math majors? e. What is the probability that at least one of the students selected is a statistics major? f. What is the probability that the selected students have different majors?

What does it mean to say that the probability that a coin toss will land head side up is \(0.5 ?\)

The National Center for Health Statistics (www.cdc .gov/nchs/data/nvsr/nvsr64/nvsr64_12.pdf, retrieved April 25,2017 ) gave the following information on births in the United States in 2014 : $$ \begin{array}{|lr|} \hline \text { Type of Birth } & \text { Number of Births } \\ \hline \text { Single birth } & 3,848,214 \\ \text { Twins } & 135,336 \\ \text { Triplets } & 4,233 \\ \text { Quadruplets } & 246 \\ \text { Quintuplets or higher } & 47 \\ \hline \end{array} $$ Use this information to estimate the probability that a randomly selected pregnant woman who gave birth in 2014 a. delivered twins b. delivered quadruplets c. gave birth to more than a single child

Roulette is a game of chance that involves spinning a wheel that is divided into 38 segments of equal size, as shown in the accompanying picture. A metal ball is tossed into the wheel as it is spinning, and the ball eventually lands in one of the 38 segments. Each segment has an associated color. Two segments are green. Half of the other 36 segments are red, and the others are black. When a balanced roulette wheel is spun, the ball is equally likely to land in any one of the 38 segments. a. When a balanced roulette wheel is spun, what is the probability that the ball lands in a red segment? b. In the roulette wheel shown, black and red segments alternate. Suppose instead that all red segments were grouped together and that all black segments were together. Does this increase the probability that the ball will land in a red segment? Explain. c. Suppose that you watch 1000 spins of a roulette wheel and note the color that results from each spin. What would be an indication that the wheel was not balanced?
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