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Consider the following information about travelers on vacation (based partly on a recent Travelocity poll): \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them, \(23 \%\) both check work email and use a cell phone to stay connected, and \(51 \%\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 88 out of every 100 who bring a laptop also check work email, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop. a. What is the probability that a randomly sclccted traveler who checks work email also uses a cell phone to stay connected? b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? c. If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/ she uses a cell phone to stay connected?

Short Answer

Expert verified
a. 0.575; b. 0.7; c. 0.7.

Step by step solution

01

Define Events

Let's define the events for this problem: \(A\) is the event that a traveler checks work email, \(B\) is the event that a traveler uses a cell phone to stay connected, and \(C\) is the event that a traveler brings a laptop. From the given data, \(P(A) = 0.4\), \(P(B) = 0.3\), \(P(C) = 0.25\), \(P(A \cap B) = 0.23\), and \(P(A^c \cap B^c \cap C^c) = 0.51\). Additionally, if a traveler brings a laptop, \(88\%\) check work email, and if a traveler uses a cell phone, \(70\%\) also bring a laptop.
02

Find Conditional Probability for Part a

For part a, we need to find the probability that a traveler uses a cell phone given that they check work email, or \(P(B|A)\). Using the formula for conditional probability, \(P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.23}{0.4} = 0.575\).
03

Find Conditional Probability for Part b

For part b, we need to find the probability that a traveler uses a cell phone given that they bring a laptop, or \(P(B|C)\). Since 70 out of every 100 who use a cell phone also bring a laptop, \(P(B|C) = 0.7\).
04

Find Combined Event Probability for Part c

For part c, we need to find the probability that a traveler uses a cell phone given that they check work email and bring a laptop, or \(P(B|A \cap C)\). Using the given data, if a traveler brings a laptop, 88 out of 100 check work email, which means \(P(A|C) = 0.88\). Therefore, the probability of checking work email and bringing a laptop is \(P(A \cap C) = P(A|C) \cdot P(C) = 0.88 \times 0.25 = 0.22\). Now, using the conditional probability formula again, \(P(B|A \cap C) = \frac{P(A \cap B \cap C)}{P(A \cap C)} = \frac{P(B \cap C) \cdot P(A|B \cap C)}{P(A \cap C)} = \frac{0.7 \cdot 0.88 \cdot 0.25}{0.22} = 0.7\). This calculation assumes independence between the email checking and phone use given laptop access, aligning with overall probabilities presented.

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics that studies uncertainty and randomness. It assigns a numerical value to the likelihood of different events occurring. These values, called probabilities, range from 0 to 1, where 0 means an event will not happen, and 1 means it definitely will.
In probability theory, we study different types of events and how they interact. Conditional probability, like what is explored in the exercise, investigates the chance of an event happening under the circumstance that another event has already occurred. This is essential in understanding how different factors influence each other in real-world scenarios, such as a traveler's likelihood of using technology based on their other actions.
When analyzing scenarios involving multiple events, the concepts of intersections (both events occur) and unions (at least one event occurs) come into play. In problems involving real life data, such as the behavior of travelers, these concepts help in indentifying patterns and predicting outcomes.
Understanding probability theory empowers individuals to make informed decisions based on potential risk and reward distribution.
Independent Events
Independent events are a crucial concept to grasp in probability calculations. Two events are considered independent if the occurrence of one event does not affect the occurrence of another. For example, if checking work email does not influence the use of a cell phone, these two events are considered independent.
However, understanding this independence is vital when dealing with conditional probabilities. If events are not independent, then the probability of one event can affect the probability of another. In the exercise, assumptions about independence needed to be carefully considered to ensure accurate calculations.
Let's look at this another way. If we say event A and event B are independent, then the probability of A and B occurring together, written as \(P(A \cap B)\), is equal to \(P(A) \cdot P(B)\). If the known probability deviates from this, it indicates dependence between the events, which means they influence each other. Recognizing whether events are independent or dependent helps create more precise models for predicting outcomes.
Probability Calculations
Probability calculations frequently employ tools like Venn diagrams and probability laws to determine how likely events are to happen. These calculations involve a variety of steps:
  • Define the events and their probabilities based on given data. For instance, in the exercise, events were defined for checking email, using a cell phone, and bringing a laptop.
  • Use conditional probability formulas, such as \(P(B|A) = \frac{P(A \cap B)}{P(A)}\). This formula helps determine the probability of one event occurring given another event has already occurred.
  • Evaluate and simplify complex probabilities by breaking them down into more manageable parts. This was evident in the solution's approach to finding probabilities of combined events.
When examining situations where multiple events interact, it's beneficial to consider how each relation and condition changes the outcome. For example, the probability of using a cell phone might change due to the use of laptops by travelers. Calculating these probabilities accurately assists in painting a clearer picture of event dynamics and supports decision-making across different scenarios.

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Most popular questions from this chapter

At a certain gas station, \(40 \%\) of the customers use regular gas \(\left(A_{1}\right), 35 \%\) use plus gas \(\left(A_{2}\right)\), and \(25 \%\) use premium \(\left(A_{3}\right)\). Of those customers using regular gas, only \(30 \%\) fill their tanks (event \(B\) ). Of those customers using plus, \(60 \%\) fill their tanks, whereas of those using premium, \(50 \%\) fill their tanks. a. What is the probability that the next customer will request plus gas and fill the tank \(\left(A_{2} \cap B\right)\) ? b. What is the probability that the next customer fills the tank? c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?

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