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A large cable company reports the following: \(80 \%\) of its customers subscribe to cable TV service \(42 \%\) of its customers subscribe to Internet service \(32 \%\) of its customers subscribe to telephone service \(25 \%\) of its customers subscribe to both cable TV and Internet service \(21 \%\) of its customers subscribe to both cable TV and phone service \(23 \%\) of its customers subscribe to both 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. In Exercise \(5.53,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. \(P(\) cable TV only) b. \(P\) (Internet \(\mid\) cable TV) c. \(P(\) exactly two services \()\) d. \(P\) (Internet and cable TV only)

Short Answer

Expert verified
The short version of the answer is: a. The probability of a customer subscribing to Cable TV only is 0.49. b. The conditional probability of a customer subscribing to Internet given they have Cable TV is 0.3125. c. The probability of a customer subscribing to exactly two services is 0.24. d. The probability of a customer subscribing to both Internet and Cable TV only is 0.1.

Step by step solution

01

Part A: Probability of Cable TV Only

To find the probability of a customer subscribing to Cable TV only, we will use the following formula: \[P(\text{Cable TV only}) = P(\text{Cable TV}) - P(\text{Cable TV and Internet}) - P(\text{Cable TV and Phone}) + P(\text{All Three})\] Now, let's substitute the given values: \(P(\text{Cable TV only}) = \frac{80}{100} - \frac{25}{100} - \frac{21}{100} + \frac{15}{100}\) \(P(\text{Cable TV only}) = \frac{49}{100}\) So, the probability of a customer subscribing to Cable TV only is 0.49.
02

Part B: Conditional Probability of Internet given Cable TV

To find the conditional probability of a customer subscribing to Internet given they have Cable TV, we will use the following formula: \[P(\text{Internet | Cable TV}) = \frac{P(\text{Cable TV and Internet})}{P(\text{Cable TV})}\] Now, let's substitute the given values: \(P(\text{Internet | Cable TV}) = \frac{\frac{25}{100}}{\frac{80}{100}}\) \(P(\text{Internet | Cable TV}) = \frac{25}{80}\) So, the conditional probability of a customer subscribing to Internet given they have Cable TV is \(\frac{25}{80}\) or 0.3125.
03

Part C: Probability of Exactly Two Services

To find the probability of a customer subscribing to exactly two services, we will add the probabilities of each pair combination without including all three services at once: \[P(\text{Exactly Two Services}) = P(\text{Cable TV and Internet only}) + P(\text{Cable TV and Phone only}) + P(\text{Internet and Phone only})\] Now, let's subtract the All Three probability from each combination: \(P(\text{Cable TV and Internet only}) = \frac{25}{100} - \frac{15}{100}\) \(P(\text{Cable TV and Phone only}) = \frac{21}{100} - \frac{15}{100}\) \(P(\text{Internet and Phone only}) = \frac{23}{100} - \frac{15}{100}\) \[P(\text{Exactly Two Services}) = \frac{10}{100} + \frac{6}{100} + \frac{8}{100}\] \(P(\text{Exactly Two Services}) = \frac{24}{100}\) So, the probability of a customer subscribing to exactly two services is 0.24.
04

Part D: Probability of Internet and Cable TV Only

To find the probability of a customer subscribing to both Internet and cable TV only, we need to subtract the probability of all three services from the probability of cable TV and Internet: \[P(\text{Internet and Cable TV only}) = P(\text{Cable TV and Internet}) - P(\text{All Three})\] Now, let's substitute the given values: \(P(\text{Internet and Cable TV only}) = \frac{25}{100} - \frac{15}{100}\) \(P(\text{Internet and Cable TV only}) = \frac{10}{100}\) So, the probability of a customer subscribing to both Internet and Cable TV only is 0.1.

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Key Concepts

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

Conditional Probability
Conditional probability helps us find the probability of an event occurring given that another event has already occurred. It is a crucial concept when dealing with interconnected events. For instance, when trying to determine the likelihood of a customer subscribing to the Internet service given they already subscribe to Cable TV. The formula for conditional probability is:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]
In this situation, \(P(A|B)\) represents the probability of event A occurring given that event B has occurred. Here, A is subscribing to the Internet and B is having Cable TV. By using this formula, one can determine how existing subscriptions (like Cable TV) affect the likelihood of other subscriptions (such as Internet).
This concept is essential in understanding the interconnectedness of events, especially in customer behavior analysis.
Probability Formulas
Probability formulas allow us to mathematically compute the likelihood of various outcomes. These formulas can address complex probabilities such as overlapping and independent events. For this exercise, some basic probability formulas are essential:
  • Addition Rule: This is used when finding the probability of either event A or event B occurring, but modifications are needed for overlapping events.
  • Subtraction Rule: When events overlap, subtract the overlapping section to avoid double-counting.
The formula for finding the probability of a single service, like Cable TV only, takes into account overlapping services:\[ P(\text{Cable TV only}) = P(\text{Cable TV}) - P(\text{Cable TV and Internet}) - P(\text{Cable TV and Phone}) + P(\text{All Three}) \]This formula effectively isolates the probability of just one service by removing overlap issues.
Hypothetical Table
A hypothetical table is a powerful tool to visually organize probabilities, especially when dealing with multiple categories or overlapping groups. This table uses assumed numbers, like 1000 customers, to break down the given percentages into tangible figures, simplifying complex percentage data into manageable parts.
  • Begin by converting percentages into actual counts based on the hypothetical total. For example, 80% of 1000 customers subscribing to Cable TV translates to 800 customers.
  • Break down overlapping sections using given percentages and fill in the corresponding cells in the table.
  • Use the table to visually separate and calculate probabilities for exclusive groups, combinations, and total possible outcomes.
This method simplifies complex probability calculations and makes it easier to see and solve problems involving multiple subscriptions. It is a highly intuitive approach often paired with theoretical formulas for clarity.

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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 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?

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 the 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 was chosen and rows corresponding to whether or not gift wrap was selected. b. Use the information in the table to calculate \(P(E \cup G)\). Give a long- run relative frequency interpretation of this probability.

Eighty-six countries won medals at the 2016 Olympics in Rio de Janeiro. Based on the results: 1 country won more than 100 medals 2 countries won between 51 and 100 medals 3 countries won between 31 and 50 medals 4 countries won between 21 and 30 medals 15 countries won between 11 and 20 medals 15 countries won between 6 and 10 medals 46 countries won between 1 and 5 medals Suppose one of the 86 countries winning medals at the 2016 Olympics is selected at random. a. What is the probability that the selected country won more than 50 medals? b. What is the probability that the selected country did not win more than 100 medals? c. What is the probability that the selected country won 10 or fewer medals? d. What is the probability that the selected country won between 11 and 50 medals?

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|cable TV) c. \(P(\) exactly two services \()\) d. \(P\) (Internet and cable TV only)

In a January 2016 Harris Poll, each of 2252 American adults was asked the following question: "If you had to choose, which ONE of the following sports would you say is your favorite?" ("Pro Football is Still America's Favorite Sport," www.theharrispoll.com/sports/Americas_Fav_Sport_2016. html, retrieved April 25,2017 ). Of the survey participants, \(33 \%\) chose pro football as their favorite sport. The report also included the following statement, "Adults with household incomes of \(\$ 75,000-<\$ 100,000(48 \%)\) are especially likely to name pro football as their favorite sport, while love of this particular game is especially low among those in \(\$ 100,000+$$ households \)(21 \%)\( Suppose that the percentages from this poll are representative of American adults in general. Consider the following events: \)F=\( event that a randomly selected American adult names pro football as his or her favorite sport \)L=\( event that a randomly selected American has a household income of \)\$ 75,000-<\$ 100,000\( \)H=\( event that a randomly selected American has a household income of \)\$ 100,000+$$ b. Are the events \(F\) and \(L\) mutually exclusive? Justify your answer. c. Are the events \(H\) and \(L\) mutually exclusive? Justify your answer. d. Are the events \(F\) and \(H\) independent? Justify your answer.

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