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Chapter 4: Problem 34

Refer to Exercise 33. Suppose that there are six prospective jurors, four men and two women, who might be chosen for the jury. Two jurors are randomly selected from these six to fill the two remaining jury seats. a. List the simple events in the experiment (HINT: There are 15 simple events if you ignore the order of selection of the two jurors.) b. What is the probability that both impaneled jurors are women?

Short Answer

Expert verified
Answer: The probability that both impaneled jurors are women is 1/15.

Step by step solution

01

List simple events in the experiment

Let's use M to represent men and W to represent women. We have 4 men (M1, M2, M3, M4) and 2 women (W1, W2). We will ignore the order of selection and list all the possible pairs: 1. (M1, M2) 2. (M1, M3) 3. (M1, M4) 4. (M2, M3) 5. (M2, M4) 6. (M3, M4) 7. (M1, W1) 8. (M1, W2) 9. (M2, W1) 10. (M2, W2) 11. (M3, W1) 12. (M3, W2) 13. (M4, W1) 14. (M4, W2) 15. (W1, W2)
02

Find the probability of selecting two women

We have a total of 15 simple events, and only one event (W1, W2) results in selecting two women. Therefore, the probability of selecting two women is: P(both impaneled jurors are women) = \frac{number\:of\:desired\:outcomes}{total\:number\:of\:outcomes} = \frac{1}{15}. Hence, the probability that both impaneled jurors are women is 1/15.

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Key Concepts

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

Understanding Simple Events
In the realm of probability, an event is deemed simple if it cannot be broken down into smaller events; it's essentially an outcome that's as uncomplicated as it gets. With the jury selection process, a simple event is a unique pairing of prospective jurors that could occur.

In our juror scenario, there are a total of 15 simple events since there are 15 unique ways to pair up the jurors, without considering the order of selection. Grasping this concept is pivotal because it lays the foundation for all probability calculations. By listing these events, we can visually and mentally ascertain all possible outcomes, establishing a comprehensive picture of the scenario before proceeding to more nuanced probability tasks.
The Fundamentals of Probability Calculations
Probability calculations are mathematical methods through which we estimate the chances of a particular event happening. In the context of our juror selection, the probability of an event can be calculated by dividing the number of ways a desired result can occur by the total number of possible outcomes.

Our textbook example illustrates this point clearly. Since we wish to find the probability of both jurors being women and there's only one simple event where this happens, we put that one desirable event (both jurors are women) over the total count of simple events (15 possible pairs), giving us a probability value of \( \frac{1}{15} \).
The Role of Combinatorics in Probability
Combinatorics is a branch of mathematics that studies counting, but not in the traditional sense of one, two, three. Instead, it involves calculating the number of ways objects can be arranged or combined, which is precisely what's needed in probability problems like our juror selection exercise.

Understanding combinatorics allows us to determine, for instance, how many ways we can pick 2 jurors out of 6 potential candidates. By utilizing combination formulas, we discover there are 15 different ways to make these pairs, resulting in our 15 simple events. This highlights how combinatorics provides the foundational figures needed for accurate probability calculations.

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Most popular questions from this chapter

A large number of adults are classified according to whether they were judged to need eyeglasses for reading and whether they actually used eyeglasses when reading. The proportions falling into the four categories are shown in the table. A single adult is selected from this group. Find the probabilities given here. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Used Eyeglasses } \\ \text { for Reading } \end{array} & \\ \hline \text { Judged to Need Eyeglasses } & \text { Yes } & \text { No } \\ \hline \text { Yes } & .44 & .14 \\ \text { No } & .02 & .40 \end{array} $$ a. The adult is judged to need eyeglasses. b. The adult needs eyeglasses for reading but does not use them. c. The adult uses eyeglasses for reading whether he or she needs them or not. d. An adult used glasses when they didn't need them.

A certain manufactured item is visually inspected by two different inspectors. When a defective item comes through the line, the probability that it gets by the first inspector is \(.1 .\) Of those that get past the first inspector, the second inspector will "miss" 5 out of \(10 .\) What fraction of the defective items get by both inspectors?

For the experiments, list the simple events in the sample space, assign probabilities to the simple events, and find the required probabilities. Three children are selected, and their gender recorded. Assume that males and females are equally likely. What is the probability that there are two boys and one girl in the group?

Five cards are selected from a 52-card deck for a poker hand. a. How many simple events are in the sample space? b. A royal flush is a hand that contains the \(\mathrm{A}, \mathrm{K}, \mathrm{Q}\), \(\mathrm{J},\) and \(10,\) all in the same suit. How many ways are there to get a royal flush? c. What is the probability of being dealt a royal flush?

A worker-operated machine produces a defective item with probability .01 if the worker follows the machine's operating instructions exactly, and with probability .03 if he does not. If the worker follows the instructions \(90 \%\) of the time, what proportion of all items produced by the machine will be defective?
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