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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

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.

N.Y. Lottery Numbers Come Up 9-1-1 on 9/11" was the headline of an article that appeared in the San Francisco Chronicle (September 13,2002 ). More than 5600 people had selected the sequence \(9-1-1\) on that date, many more than is typical for that sequence. A professor at the University of Buffalo is quoted as saying, "I'm a bit surprised, but I wouldn't characterize it as bizarre. It's randomness. Every number has the same chance of coming up." a. The New York state lottery uses balls numbered \(0-9\) circulating in three separate bins. To select the winning sequence, one ball is chosen at random from each bin. What is the probability that the sequence \(9-1-1\) is the sequence selected on any particular day? (Hint: It may be helpful to think about the chosen sequence as a threedigit number.) b. What approach (classical, relative frequency, or subjective) did you use to obtain the probability in Part (a)? Explain.

An individual is presented with three different glasses of cola, labeled C, D, and P. He is asked to taste all three and then list them in order of preference. Suppose that the same cola has actually been put into all three glasses. a. What are the simple events in this chance experiment, and what probability would you assign to each one? b. What is the probability that \(\mathrm{C}\) is ranked first? c. What is the probability that \(\mathrm{C}\) is ranked first \(a n d \mathrm{D}\) is ranked last?

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let not \(E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(\left.E_{1}\right)\) ? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

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.)
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