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Chapter 6: Problem 30

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7\), \(P(F)=.6\), and \(P(E \cap F)=.54 .\) a. Calculate \(P(E \mid F)\), the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

Short Answer

Expert verified
The probability that the first airline's flight is fully booked given that the second airline's flight is fully booked (\(P(E | F)\)) is 0.9. The probability that the second airline's flight is fully booked given that the first airline's flight is fully booked (\(P(F | E)\)) is 0.771.

Step by step solution

01

Understanding conditional probability

Before diving into calculations, it's essential to understand the concept of conditional probability. In general, the conditional probability of an event \(A\) given that another event \(B\) has occurred is calculated as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]Where \(P(A \cap B)\) is the probability that both events \(A\) and \(B\) occur, and P(B) is the probability that event \(B\) occurs.
02

Calculate \(P(E | F)\)

To solve for \(P(E | F)\), let's plug the given values into the formula:\[ P(E | F) = \frac{P(E \cap F)}{P(F)} \]Using the given values,\[ P(E | F) = \frac{.54}{.6} \]
03

Simplify the expression

Solving this fraction will give us the value of \(P(E | F)\).
04

Calculate \(P(F | E)\)

To solve for \(P(F | E)\), we use the same formula as in Step 2, but this time calculating the probability that event F occurs, given that event E has occurred:\[ P(F | E) = \frac{P(E \cap F)}{P(E)} \]Using the given values,\[ P(F | E) = \frac{.54}{.7} \]
05

Simplify the expression

Solving this fraction will give us the value of \(P(F | E)\).

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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 concept that evaluates the probability of two or more events occurring simultaneously. Let's imagine two events happening in sequence, each with its own chance of happening. Consider an example like flipping a coin and rolling a die. You might want to find the chance of getting heads and rolling a six at the same time. That's where joint probability comes in handy.

In the context of our exercise, joint probability is illustrated by the event where both airlines' flights are fully booked on the same day. Symbolically, it is represented as \(P(E \cap F)\). This represents the likelihood that both event \(E\) (the first airline's flight is fully booked) and event \(F\) (the second airline's flight is fully booked) occur at the same time.

To calculate joint probability:
  • Use known probabilities of individual events
  • Understand that joint probability assumes interdependency between events
In problems involving conditional probability, joint probability can provide a base for understanding relationships between dependent events.
Independent Events
Independent events are those whose occurrence doesn't affect each other. It's like rolling two different dice without them influencing each other's outcomes. If two events are independent, the probability of both occurring is the product of their individual probabilities.

For our flight exercise, events \(E\) and \(F\) are not independent because the occurrence probability of one flight being fully booked can influence the other. To check for independence, see if the product of their individual probabilities equates to their joint probability. For independent events, the formula is:\[ P(E) \times P(F) = P(E \cap F) \]If equality holds, then events are independent. Here, you would calculate:\[ 0.7 \times 0.6 = 0.42 \]But since \(P(E \cap F) = 0.54\), the flights being fully booked seems to be dependent on each other, as the probabilities differ.
Probability Theory
Probability theory is the mathematical framework that helps us measure uncertainty and determine likelihoods for events happening. It's like having a toolkit to assess risk in various scenarios.

In probability theory, we work with concepts such as conditional probability, which helps us adjust probabilities based on new information. This concept was key in our exercise, allowing us to figure out \(P(E | F)\) and \(P(F | E)\), the probabilities of one flight being fully booked if you know the status of the other.

Fundamental principles of probability include:
  • Probabilities are between 0 and 1, depicting impossibility to certainty.
  • Sum of probabilities for all possible outcomes of a particular event equals 1.
  • Conditional probability involves using known information to adjust likelihoods.
Understanding these principles is crucial. They ensure that you can compute probabilities accurately, whether it's for joint happenings or interdependent events like our airline flights.

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Most popular questions from this chapter

Three friends \((A, B\), and \(C)\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\mathrm{~A}\) beats \(\mathrm{B})=.7, P(\mathrm{~A}\) beats C) \(=.8, P(\mathrm{~B}\) beats \(\mathrm{C})=.6\), and that the outcomes of the three matches are independent of one another. a. What is the probability that \(\mathrm{A}\) wins both her matches and that \(\mathrm{B}\) beats \(\mathrm{C}\) ? b. What is the probability that \(A\) wins both her matches? c. What is the probability that \(A\) loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

The Australian newspaper The Mercury (May 30 , 1995) reported that, based on a survey of 600 reformed and current smokers, \(11.3 \%\) of those who had attempted to quit smoking in the previous 2 years had used a nicotine aid (such as a nicotine patch). It also reported that \(62 \%\) of those who quit smoking without a nicotine aid began smoking again within 2 weeks and \(60 \%\) of those who used a nicotine aid began smoking again within 2 weeks. If a smoker who is trying to quit smoking is selected at random, are the events selected smoker who is trying to quit uses a nicotine aid and selected smoker who bas attempted to quit begins smoking again within 2 weeks independent or dependent events? Justify your answer using the given information.

Consider the following information about travelers on vacation: \(40 \%\) check work e-mail, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) both check work e-mail and use a cell phone to stay connected, and \(51 \%\) neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. In addition \(88 \%\) of those who bring a laptop also check work e-mail and \(70 \%\) of those who use a cell phone to stay connected also bring a laptop. With \(E=\) event that a traveler on vacation checks work e-mail, \(C=\) event that a traveler on vacation uses a cell phone to stay connected, and \(L=\) event that a traveler on vacation brought a laptop, use the given information to determine the following probabilities. A Venn diagram may help. a. \(P(E)\) b. \(P(C)\) c. \(P(L)\) d. \(P(E\) and \(C)\) e. \(P\left(E^{C}\right.\) and \(C^{C}\) and \(\left.L^{C}\right)\) f. \(P(E\) and \(\operatorname{Cor} L)\) g. \(\quad P(E \mid L)\) h. \(P(L \mid C)\) i. \(P(E\) and \(C\) and \(L)\) j. \(P(E\) and \(L)\) k. \(P(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a schedule is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is \(.8\). 2\. If Maria completes her part on time, the probability that Alex completes on time is \(.9\), but if Maria is late, the probability that Alex completes on time is only \(.6 .\) 3\. If Alex completes his part on time, the probability that Juan completes on time is \(.8\), but if Alex is late, the probability that Juan completes on time is only \(.5 .\) 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(.9\), but if Juan is late, the probability that Jacob completes on time is only .7. Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.

The article "Chances Are You Know Someone with a Tattoo, and He's Not a Sailor" (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50 . The accompanying data are consistent with summary values given in the article. $$ \begin{array}{l|cc} & \text { At Least One Tattoo } & \text { No Tattoo } \\ \hline \text { Age 18-29 } & 18 & 32 \\ \text { Age 30-50 } & 6 & 44 \\ \hline \end{array} $$ Assuming these data are representative of adult Americans and that an adult American is selected at random, use the given information to estimate the following probabilities. a. \(P(\) tattoo \()\) b. \(P(\) tattoo \(\mid\) age \(18-29)\) c. \(P(\) tattoo \(\mid\) age \(30-50\) ) d. \(P(\) age \(18-29 \mid\) tattoo \()\)
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