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Chapter 6: Problem 59

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 \(P(F \mid E) .)\)

Short Answer

Expert verified
The probabilities are: \(P(E)\) is 0.6, \(P(F | E)\) is \(\frac{5}{9}\), and \(P(E \: and \: F)\) is \(\frac{1}{3}\).

Step by step solution

01

Calculate \(P(E)\)

The event \(E\) is that the first vehicle assigned is a van. Since there are 6 vans out of 10 vehicles, the probability that a randomly chosen vehicle is a van is given by \(\frac{number \:of \:vans}{total\: number\: of\: vehicles} = \frac{6}{10} = 0.6\). So, \(P(E) = 0.6\).
02

Calculate \(P(F | E)\)

The event \(F\) is that the second vehicle assigned is a van. \(P(F | E)\) is the probability that the second vehicle assigned is a van given that the first one already assigned was a van. After assigning one van, there are now 5 vans and 9 vehicles left. So, \(P(F | E) = \frac{number\: of\: remaining\: vans}{total\: number\: of\: remaining\: vehicles} = \frac{5}{9}.\) So, \(P(F | E) = \frac{5}{9}\).
03

Calculate \(P(E \: and \:F)\)

According to the definition of conditional probability, the probability that both events E and F occur, which is denoted as \(P(E \: and \: F)\), is the product of the probability of E and the probability of F given E. So, \(P(E \: and \: F) = P(E) \times P(F | E) = 0.6 \times \frac{5}{9} = \frac{1}{3}. \) So, \(P(E \: and \: F) = \frac{1}{3}\).

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Key Concepts

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

Conditional Probability
In statistics, one of the most fundamental concepts is that of conditional probability, which measures the probability of an event occurring given that another event has already happened. The conditional probability can be represented as P(A|B), which reads as 'the probability of A given B'.

To understand this with an example, let's consider the textbook exercise which deals with the assignment of vehicles. If the event E is that the first vehicle assigned is a van, and we want to find the conditional probability that the second vehicle assigned is also a van (event F) given that E has already occurred, we use the formula:
\[P(F | E) = \frac{P(E \: and \: F)}{P(E)}\]
In the solution provided, after one van has already been assigned (event E), there are fewer vans to choose from for the second assignment. The conditional probability P(F|E) takes into account this reduced pool of vehicles. This concept helps in understanding probabilities in more complex, dependent scenarios and is widely used in fields ranging from games of chance to predictive modelling.
Random Assignment
Another important concept in probability is random assignment, which refers to the practice of assigning subjects or objects to groups in a way that each subject has an equal chance of being placed in any group. This principle is crucial for conducting fair and unbiased experiments or selections. In the context of our example, the random assignment principle is applied when choosing vehicles to fulfill requests.

The term 'completely random fashion' as stated in the exercise implies each of the 10 vehicles, which includes 6 vans and 4 cars, has an equal probability of being selected. Random assignment ensures that the selection process doesn't favor vans over cars, or vice versa. It's the randomness that maintains the equality of chances among all vehicles, which is fundamental for calculating event probabilities accurately.
Event Probability
Finally, the concept of event probability signifies the chance of an event occurring. It is a measure that varies between 0 and 1, with 0 indicating an impossibility and 1 indicating a certainty. In our example, the event probability is first calculated for the event that a van is selected as the first vehicle:
\[P(E) = \frac{6}{10} = 0.6\]
This outcome represents the likelihood of event E happening in a single trial.

When the first event's outcome influences the second event, the calculation must consider the changed conditions, as seen when calculating P(F|E). The probability of both events E and F occurring is found by multiplying the probability of E with the conditional probability P(F|E).
Enhancing the students' understanding of these probabilities helps in grasping the underlying mechanism of random processes and is pivotal for studies in fields that range from finance to engineering.

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Most popular questions from this chapter

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.)

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 \(L)\)

A company that manufactures video cameras produces a basic model and a deluxe model. Over the past year, \(40 \%\) of the cameras sold have been the basic model. Of those buying the basic model, \(30 \%\) purchase an extended warranty, whereas \(50 \%\) of all purchasers of the deluxe model buy an extended warranty. If you learn that a randomly selected purchaser bought an extended warranty, what is the probability that he or she has a basic model?

Of the 60 movies reviewed last year by two critics on their joint television show, Critic 1 gave a "thumbs-up" rating to 15 , Critic 2 gave this rating to 20 , and 10 of the movies were rated thumbs-up by both critics. Suppose that 1 of these 60 movies is randomly selected. a. Given that the movie was rated thumbs-up by Critic 1 , what is the probability that it also received this rating from Critic \(2 ?\) b. If the movie did not receive a thumbs-up rating from Critic 2, what is the probability that it also did not receive a thumbs up rating from Critic \(1 ?\) (Hint: Construct a table with two rows for the first critic [for "up" and "not up"] and two columns for the second critic: then enter the relevant probabilities.)

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: a. The probability that grip size is \(4 \frac{1}{2}\) in. (event \(\mathrm{A}\) ) is $$ P(A)=P\left(O_{2} \text { or } O_{5}\right)=.20+.15=.35 $$ How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) in. c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) in.?
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