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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. (Hint: See Example 5.6\()\) 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

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a) The '1000 customer' table has 250 customers subscribing to both services, 550 subscribing to cable TV only and 170 subscribing to the Internet only. b) i) The probability that a customer subscribes to both services is 25% and ii) the probability that a customer subscribes to exactly one service is 72%.

Step by step solution

01

Construct the hypothetical 1000 table

Assume there are 1000 customers. The number that subscribe to the cable TV service is \(80 \% \times 1000 = 800\), the number that subscribe to the Internet service is \(42 \% \times 1000 = 420\), and the number that subscribe to either one of these services is \(97 \% \times 1000 = 970\). From this, it can be found that the number of customers who subscribe to both services is \(800 + 420 - 970 = 250\). Therefore, 250 customers subscribe to both, 550 subscribe to cable TV only and 170 subscribe to Internet only.
02

Find the probability for a customer subscribing to both services

The probability that a randomly selected customer subscribes to both services can be calculated by dividing the number of customers subscribing to both services by the total number of customers: \( \frac{250}{1000} = 0.25 \) or 25 \%.
03

Find the probability for a customer subscribing to exactly one service

The probability that a randomly selected customer subscribes to exactly one service can be calculated by adding the probabilities of subscribing to only one of the two services: \( \frac{550 (TV only) + 170 (Internet only)}{1000 (total)} = 0.72 \) or 72\%.

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Key Concepts

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

Hypothetical 1000 Table
The concept of a Hypothetical 1000 Table is a useful tool in probability and statistics, helping us visualize data in terms of manageable numbers. Imagine you have 1000 hypothetical customers, and this helps simplify percentage calculations. For example, if 80% of customers subscribe to cable TV, that translates to 800 customers in this hypothetical scenario. Similarly, 42% for Internet translates to 420 customers. This method makes probabilities easier to understand and calculate.
It also becomes simpler to deal with overlapping probabilities. We were told that 97% subscribe to at least one service. The overlap, or both subscribing to cable and Internet, can be calculated by adding the individual numbers and subtracting those accounted twice (in both categories). Doing so gives us 250 customers or 25% of the sample subscribing to both services. This table break down can swiftly be used to calculate other required probabilities. It turns abstract percentages into concrete numbers aiding in better comprehension.
Joint Probability
Joint Probability refers to the probability of two events happening at the same time. In the cable company scenario, this means a customer subscribes to both cable TV and the Internet. We calculate joint probability by dividing the number of customers subscribing to both services by the total number of customers.
Joint probability is a key concept, especially when dealing with dependent events. Here, the joint probability is calculated as:
  • Number of customers subscribing to both services = 250
  • Total customers = 1000
This leads to a joint probability of \(\frac{250}{1000} = 0.25\) or 25%. Understanding joint probability is critical in statistics to analyze how two events correlate with each other and their combined likelihood in any given population.
Mutually Exclusive Events
Events are considered mutually exclusive if they cannot happen at the same time. For example, in our case, subscribing to only cable TV and only Internet are mutually exclusive events, because a customer can only subscribe to one in this context, not both concurrently.
Knowing which events are mutually exclusive helps in determining probabilities, especially when determining 'either-or' scenarios. For instance, the probability of a customer subscribing to exactly one service involved simply adding those subscribing exclusively to either TV or Internet:
  • Cable TV only = 550 customers
  • Internet only = 170 customers
Thus, the probability of subscribing to exactly one service is \[\frac{550 + 170}{1000} = 0.72\] or 72%. This is because both events don't overlap; one can observe how the probabilities are additive for mutually exclusive events.

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Most popular questions from this chapter

Roulette is a game of chance that involves spinning a wheel that is divided into 38 equal segments, 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?

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 \(P(E \cap F)=0.15 .\) (Hint: See Example 5.5) a. Use the given probability information to set up a "hypothetical \(1000 "\) table with columns corresponding to \(E\) and not \(E\) and rows corresponding to \(F\) and not \(F\). b. Use the table from Part (a) to find the following probabilities: i. the probability that Shelly must stop for at least one light (the probability of \(E \cup F)\). ii. the probability that Shelly does not have to stop at either light. iii. the probability that Shelly must stop at exactly one of the two lights. iv. the probability that Shelly must stop only at the first light.

A large cable TV company reports the following: \- \(80 \%\) of its customers subscribe to its cable TV service \- \(42 \%\) of its customers subscribe to its Internet service \- \(32 \%\) of its customers subscribe to its telephone service \(25 \%\) of its customers subscribe to both its cable TV and Internet service \(21 \%\) of its customers subscribe to both its cable TV and phone service \- \(23 \%\) of its customers subscribe to both its Internet and phone service \- \(15 \%\) of its customers subscribe to all three services Consider the chance experiment that consists of selecting one of the cable company customers at random. Find and interpret the following probabilities: a. \(P(\) cable TV only \()\) b. \(P(\) Internet \(\mid\) cable \(\mathrm{TV})\) c. \(P\) (exactly two services) d. \(P\) (Internet and cable TV only)

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.

An online store offers two methods of shipping-regular ground service and an expedited 2 -day shipping. Customers may also choose whether or not to have a purchase gift wrapped. Suppose that the events \(E=\) event that the customer chooses expedited shipping \(G=\) event that the customer chooses gift wrap are independent with \(P(E)=0.26\) and \(P(G)=0.12\). a. Construct a "hypothetical 1000 " table with columns corresponding to whether or not expedited shipping is chosen and rows corresponding to whether or not gift wrap is selected. b. Use the table to calculate \(P(E \cup G)\). Give a long-run relative frequency interpretation of this probability.

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