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Chapter 5: Problem 89

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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We assumed there are 1000 customers in total. In our hypothetical 1000 table, we found that 230 customers have both Internet and Phone service, 190 have Internet service only, and 90 have Phone service only. We calculated the following probabilities: a) Probability of a randomly selected customer subscribing to both Internet and Phone service is \(0.23\) or \(23\%\). b) Probability of a randomly selected customer subscribing to exactly one of the two services is \(0.28\) or \(28\%\).

Step by step solution

01

Determine total number of customers

We are given probabilities, but we need to work with actual customer counts in order to set up a hypothetical 1000 table. So let's assume there are 1000 customers in total.
02

Calculate counts for each service

We are given the following probabilities: - Internet service: \(42\% \) - Phone service: \(32 \% \) - Internet or Phone service (or both): \(51 \% \) Now, we will convert these probabilities into counts for our 1000 table. - Internet service: \(1000 \times 0.42 = 420\) - Phone service: \(1000 \times 0.32 = 320\) - Internet or Phone service (or both): \(1000 \times 0.51 = 510\)
03

Calculate count for both services

To find out how many customers subscribe to both services, we can use the formula for the union of two sets: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) In our case, we have: - \(P(A \cap B) = P(\textrm{Internet and Phone service}) \) - \(P(A) =\) 420 customers (Internet service) - \(P(B) =\) 320 customers (Phone service) - \(P(A \cup B) =\) 510 customers (Internet or Phone service or both) We can now solve for \(P(A \cap B)\): \(P(A \cap B) = 420 + 320 - 510 = 230\) So, 230 customers have both services. Our hypothetical 1000 table will look like: | | Internet | Not Internet | |----------------|----------|---------------| | Phone | **230** | 90 | | Not Phone | 190 | 500 | Now, let's move on to finding the probabilities that were asked for. #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.#
04

Calculate the probability

To find the probability that a randomly selected customer subscribes to both the Internet service and the phone service, we can use the formula: \(P(A \cap B) = \dfrac{\text{Number of customers that have both services}}{\text{Total number of customers}}\) So, we have: \(P(A \cap B) = \dfrac{230}{1000} = 0.23\) So the probability that a randomly selected customer subscribes to both the Internet service and the phone service is 0.23 or 23%. #ii. the probability that a randomly selected customer subscribes to exactly one of the two services.#
05

Calculate the probability

To find the probability that a randomly selected customer subscribes to exactly one of the two services, we can use the following formula: \(P(\textrm{Exactly one service}) = P(A \textrm{ only}) + P(B \textrm{ only})\) From our 1000 table, we have: - \(P(A \textrm{ only}) =\dfrac{190}{1000}\) (Internet service only) - \(P(B \textrm{ only}) =\dfrac{90}{1000}\) (Phone service only) So, we have: \(P(\textrm{Exactly one service}) = \dfrac{190}{1000} + \dfrac{90}{1000} = \dfrac{280}{1000} = 0.28\) So the probability that a randomly selected customer subscribes to exactly one of the two services is 0.28 or 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 be complex, but the 'hypothetical 1000 table' simplifies matters. When percentages are given, envisioning a total of 1000 helps in drawing a clear picture of the data. Why 1000? Because it eases the conversion of percentages into actual numbers that can be more readily visualized.

For instance, if 42% of customers use internet service, imagine 420 out of 1000 customers. Similarly, for phone services at 32%, visualize 320 customers. A hypothetical 1000 table aids in arranging these figures for clarity and, crucially, helps in calculating intersections and unions of sets, fundamental concepts in probability theory.

Using a hypothetical 1000 table can notably assist students struggling to grasp abstract probability concepts, making the math more tangible. For educators, integrating this strategy into explanations or teaching frameworks could be highly beneficial in demystifying probability for their students.
Union of Sets
In the realm of probability theory, the 'union of sets' refers to the combination of two or more sets. Think of each set as a basket of unique elements. When we take the union of two sets, we're essentially combining all elements from both baskets while removing duplicates.

The probability of the union of two sets, written as \(P(A \cup B)\), doesn't simply add the individual probabilities鈥攖his could lead to counting shared elements twice. To accurately calculate the union, subtract the probability of the intersection (shared elements) from the sum of the individual probabilities. The formula is \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

For educators teaching probability, it's vital to emphasize the importance of not double-counting. Exercises that allow students to practice with real data can help solidify their understanding of unions in probability.
Probability of Intersection
Intersecting sets and the probability of their intersection, denoted as \(P(A \cap B)\), is another cornerstone of probability theory. The intersection involves elements that belong to both sets simultaneously.

Consider a Venn diagram with two overlapping circles. The overlapping section is the intersection, containing elements common to both circles. In probability, the intersection represents outcomes that satisfy both conditions associated with sets A and B.

Correctly identifying shared elements between sets is crucial: it prevents overestimation of probabilities and helps in understanding dependent and independent events. A good way to master calculation of intersections is through consistent practice with a variety of scenarios, as repeated exposure to these concepts solidifies comprehension and fosters mathematical intuition.

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Most popular questions from this chapter

If you were to roll a fair die 1000 times, about how many sixes do you think you would observe? What is the probability of observing a six when a fair die is rolled?

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