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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

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.?

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.

At a large university, the Statistics Department has tried a different text during each of the last three quarters. During the fall quarter, 500 students used a book by Professor Mean; during the winter quarter, 300 students used a book by Professor Median; and during the spring quarter, 200 students used a book by Professor Mode. A survey at the end of each quarter showed that 200 students were satisfied with the text in the fall quarter, 150 in the winter quarter, and 160 in the spring quarter. a. If a student who took statistics during one of these three quarters is selected at random, what is the probability that the student was satisfied with the textbook? b. If a randomly selected student reports being satisfied with the book, is the student most likely to have used the book by Mean, Median, or Mode? Who is the least likely author? (Hint: Use Bayes' rule to compute three probabilities.)

The newspaper article "Folic Acid Might Reduce Risk of Down Syndrome" (USA Today, September 29 , 1999) makes the following statement: "Older women are at a greater risk of giving birth to a baby with Down Syndrome than are younger women. But younger women are more fertile, so most children with Down Syndrome are born to mothers under \(30 .\) " Let \(D=\) event that a randomly selected baby is born with Down Syndrome and \(Y=\) event that a randomly selected baby is born to a young mother (under age 30 ). For each of the following probability statements, indicate whether the statement is consistent with the quote from the article, and if not, explain why not. a. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) b. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.7\) c. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) d. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\) e. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.4\) f. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\)

Suppose that a new Internet company Mumble.com requires all employees to take a drug test. Mumble.com can afford only the inexpensive drug test - the one with a \(5 \%\) false-positive rate and a \(10 \%\) false-negative rate. (That means that \(5 \%\) of those who are not using drugs will incorrectly test positive and that \(10 \%\) of those who are actually using drugs will test negative.) Suppose that \(10 \%\) of those who work for Mumble.com are using the drugs for which Mumble is checking. (Hint: It may be helpful to draw a tree diagram to answer the questions that follow.) a. If one employee is chosen at random, what is the probability that the employee both uses drugs and tests positive? b. If one employee is chosen at random, what is the probability that the employee does not use drugs but tests positive anyway? c. If one employee is chosen at random, what is the probability that the employee tests positive? d. If we know that a randomly chosen employee has tested positive, what is the probability that he or she uses drugs?
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