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Chapter 6: Problem 28

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 is 0.9. The probability that the second airline's flight is fully booked given that the first airline's flight is fully booked is 0.7714.

Step by step solution

01

Define Notation

Start by defining the probability of each event and the intersection of these events. Here, \(P(E)=0.7\) is the probability of the first airline's flight being fully booked, \(P(F)=0.6\) is the probability of the second airline's flight being fully booked and \(P(E 鈭� F)=0.54\) is the probability of both flights being fully booked.
02

Calculate \(P(E \mid F)\)

To calculate the conditional probability, we need to apply the formula \(P(E \mid F) = P(E 鈭� F) / P(F)\). Inserting given probabilities, we get \(P(E \mid F) = 0.54 / 0.6 = 0.9\). Thus, the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked is 0.9.
03

Calculate \(P(F \mid E)\)

Now, we calculate \(P(F \mid E)\) the probability that the second airline's flight is fully booked given that the first airline's flight is fully booked, following a similar method. We use the formula, \(P(F \mid E) = P(E 鈭� F) / P(E)\), yielding: \(P(F \mid E) = 0.54 / 0.7 = 0.7714\). Thus, the probability that the second airline's flight is fully booked given that the first airline's flight is fully booked is 0.7714.

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Key Concepts

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

Intersection of Events
In probability, the intersection of events describes the scenario where two or more events happen simultaneously. Imagine you're analyzing two flights from different airlines, like in our exercise. One is fully booked (event E), and the other is also fully booked (event F). The probability of both these events occurring together at the same time on a particular day is expressed as \(P(E \cap F)\).
To better understand this concept:
  • The intersection focuses on joint outcomes 鈥� here, both flights being fully booked.
  • If events E and F are independent, the occurrence of one does not influence the other. However, we often deal with dependent events in practice, where knowing one event happened changes the probability of the other.
  • In our problem, \(P(E \cap F) = 0.54\). That's the chance both airlines鈥� flights are fully booked on the same day.
Understanding intersections helps us when we delve into more complex probability calculations, like conditional probability.
Probability of Events
Probability helps us measure the chance of a specific outcome occurring. In our exercise, we're dealing with simple events and their probabilities. For instance, \(P(E) = 0.7\) reflects the likelihood of the first airline's flight being fully booked. Similarly, \(P(F) = 0.6\) is the likelihood of the second airline鈥檚 flight meeting the same fate.
Some key points to consider about probabilities of events:
  • Each event's probability ranges from 0 to 1, where 0 indicates no chance and 1 means certainty.
  • For mutually exclusive events, however, their intersection (or joint probability) is zero.
  • But in our case, the events are not mutually exclusive because there exists a non-zero probability (\(P(E \cap F)\)) where both flights are fully booked.
Using these basic probabilities sets the stage for understanding how events interact, especially when conditioned on each other.
Conditional Probability Formula
The Conditional Probability Formula allows us to compute the probability of one event given the occurrence of another. This is especially useful when we know two events may interact or be dependent on each other.The formula is expressed as:\[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\]This formula calculates the probability of event A happening, given event B has happened. In the exercise, we calculated:
  • \(P(E \mid F)\), where E is the first airline鈥檚 flight being fully booked, given the second flight (event F) is fully booked:
  • Inserting values gives \(P(E \mid F) = 0.54 / 0.6 = 0.9\). This means there's a 90% chance that the first flight is fully booked if the second is as well.
  • Similarly, for \(P(F \mid E)\), we substitute the respective values: \(P(F \mid E) = 0.54 / 0.7 = 0.7714\). Meaning, if the first flight is fully booked, the second has a 77.14% chance of being fully booked too.
Understanding the conditional probability formula allows for complex predictive insights in everyday situations, such as forecast planning based on known occurrences.

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Most popular questions from this chapter

A certain university has 10 vehicles available for use by faculty and staff. Six of these are vans and four are cars. On a particular day, only two requests for vehicles have been made. Suppose that the two vehicles to be assigned are chosen in a completely random fashion from among the 10 . a. Let \(E\) denote the event that the first vehicle assigned is a van. What is \(P(E)\) ? b. Let \(F\) denote the probability that the second vehicle assigned is a van. What is \(P(F \mid E)\) ? c. Use the results of Parts (a) and (b) to calculate \(P(E\) and \(F)\) (Hint: Use the definition of \(P(F \mid E) .)\)

Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let \(\mathrm{D}=\) defective and \(\mathrm{N}=\) not defective). The chance experiment terminates as soon as a nondefective battery is obtained. a. Give five possible experimental outcomes. b. What can be said about the number of outcomes in the sample space? c. What outcomes are in the event \(E\), that the number of batteries examined is an even number?

A mutual fund company offers its customers several different funds: a money market fund, three different bond funds, two stock funds, and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: \(\begin{array}{lr}\text { Money market } & 20 \% \\ \text { Short-term bond } & 15 \% \\ \text { Intermediate-term bond } & 10 \% \\ \text { Long-term bond } & 5 \% \\ \text { High-risk stock } & 18 \% \\ \text { Moderate-risk stock } & 25 \% \\ \text { Balanced fund } & 7 \%\end{array}\) A customer who owns shares in just one fund is to be selected at random. a. What is the probability that the selected individual owns shares in the balanced fund? b. What is the probability that the individual owns shares in a bond fund? c. What is the probability that the selected individual does not own shares in a stock fund?

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books at random to the four stu- dents \((1,2,3\), and 4\()\) who claim to have left books. One possible outcome is that 1 receives 2 's book, 2 receives 4 's book, 3 receives his or her own book, and 4 receives 1 's book. This outcome can be abbreviated \((2,4,3,1)\). a. List the 23 other possible outcomes. b. Which outcomes are contained in the event that exactly two of the books are returned to their correct owners? As- suming equally likely outcomes, what is the probability of this event? c. What is the probability that exactly one of the four students receives his or her own book? d. What is the probability that exactly three receive their own books? e. What is the probability that at least two of the four students receive their own books?

Consider the following information about travelers on vacation: \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) 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 \(88 \%\) of those who bring a laptop also check work email 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 email, \(C=\) 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 \(L^{C}\) ) f. \(P(\) Eor C or \(L\) ) g. \(P(E \mid L)\) j. \(P(E\) and \(L)\) h. \(P(L \mid C)\) k. \(P(C\) and \(L)\) i. \(P(E\) and \(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)
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