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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

Refer to Exercise 6.18. Adding probabilities in the first row of the given table yields \(P(\) midsize \()=.45\), whereas from the first column, \(\mathrm{P}\left(4 \frac{3}{8}\right.\) in. grip) \(=.30\). Is the following true? $$ P\left(\text { midsize } \text { or } 4 \frac{3}{8} \text { in. grip }\right)=.45+.30=.75 $$ Explain.

Components of a certain type are shipped to a supplier in batches of \(10 .\) Suppose that \(50 \%\) of all batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. A batch is selected at random. Two components from this batch are randomly selected and tested. a. If the batch from which the components were selected actually contains two defective components, what is the probability that neither of these is selected for testing? b. What is the probability that the batch contains two defective components and that neither of these is selected for testing? c. What is the probability that neither component selected for testing is defective? (Hint: This could happen with any one of the three types of batches. A tree diagram might help.)

6.12 Consider a Venn diagram picturing two events \(A\) and \(B\) that are not disjoint. a. Shade the event \((A \cup B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C} .\) How are these two events related? b. Shade the event \((A \cap B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C}\). How are these two events related? (Note: These two relationships together are called DeMorgan's laws.)

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\) and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F ?\) b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F) ?\) A Venn diagram might help.)

A deck of 52 cards is mixed well, and 5 cards are dealt. a. It can be shown that (disregarding the order in which the cards are dealt) there are \(2,598,960\) possible hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit? b. It can be shown that exactly 63,206 hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented? c. Using the result of Part (b), what is the probability that a hand contains cards from exactly two suits?
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