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Chapter 4: Problem 11

Three people are randomly selected to report for jury duty. The gender of each person is noted by the county clerk. a. Define the experiment. b. List the simple events in \(S\). c. If each person is just as likely to be a man as a woman, what probability do you assign to each simple event? d. What is the probability that only one of the three is a man? e. What is the probability that all three are women?

Short Answer

Expert verified
Answer: The probability of having only one man in the group of three is \(\frac{3}{8}\).

Step by step solution

01

Experiment Definition

The experiment involves randomly selecting three people for jury duty and noting the gender of each person. #b. List the simple events in S#
02

Simple Events

There are two gender possibilities for each person (man or woman) and three people. Therefore, there are \(2^3 = 8\) simple events in the sample space, S. They are: (MMM), (MMW), (MWM), (MWW), (WMM), (WMW), (WWM), and (WWW), where M denotes a man and W denotes a woman. #c. Probability assignment to simple events#
03

Probability Assignment

Since each person is equally likely to be a man or a woman, the probability of each simple event is equal. That is, \(\frac{1}{8}\). #d. Probability that only one of the three is a man#
04

Probability of One Man

To find the probability of having only one man in the group of three, we need to count the simple events that have only one man. These events are (MWW), (WMW), and (WWM). So, 3 out of 8 events meet this condition. Thus, the probability is \(\frac{3}{8}\). #e. Probability that all three are women#
05

Probability of All Women

To find the probability that all three are women, we need to find the simple event that has all women. This event is (WWW). There is only 1 event with all women. Thus, the probability is \(\frac{1}{8}\).

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Key Concepts

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

Simple Events
In probability, a simple event is a specific outcome or a single possibility of what might happen in an experiment. For example, imagine you flip a coin; your simple events are 'heads' or 'tails'. In the scenario of jury duty selection, the simple events are different sequences of gender outcomes for three individuals. Each sequence like (MMM), indicating all men, or (MWW), indicating various combinations of man and woman, represents a simple event.

Here are some points to understand simple events better:
  • Each simple event in a probability experiment is mutually exclusive.
  • All potential simple events form what is known as the sample space.
  • In the given problem, there are 8 simple events for 3 people, with possible sequences being (MMM, MMW, MWM, MWW, WMM, WMW, WWM, WWW).
Understanding simple events is crucial, as these form the foundational basis for calculating probabilities.
Sample Space
The sample space in probability is the total collection of all possible outcomes or results of an experiment. It's like the "list" where all individual simple events live. For the jury duty scenario, the sample space encompasses all possible gender combinations of three individuals.

The sample space is expressed as: \[ S = \{ (MMM), (MMW), (MWM), (MWW), (WMM), (WMW), (WWM), (WWW) \} \]

Some points to remember about a sample space:
  • It includes every single outcome that can happen in the experiment.
  • The sample space should be exhaustive, covering all possibilities.
  • The probability of each simple event in the sample space must add up to 1.
This concept helps in visualizing all possible outcomes and is essential for determining probabilities of events.
Gender Probability
Gender probability in this context refers to the likelihood of selecting a man or a woman for the jury. Since each individual is equally likely to be either gender, each person has a probability of \( \frac{1}{2} \).

Given the scenario, probability derives from:
  • Each simple event in the sample space having uniform probability.
  • For three people with two gender options each (man or woman), probability for each simple event is \( \frac{1}{8} \), considering there are 8 possible outcomes.
This uniform probability distribution allows us to calculate the likelihood of specific events, such as all members being women (WWW) or having exactly one man (e.g., MWW). Therefore, understanding gender probability is vital for calculating various combinations of outcomes.
Combinatorics
Combinatorics is the field of mathematics concerned with counting, combinations, and permutations. It provides techniques to calculate the number of ways an event can occur. In our situation, combinatorics helps to determine possible gender combinations of three people.

Some vital concepts in combinatorics include:
  • Permutations: Arrangements where order matters. For example, with three different people arranging their gender into slots.
  • Combinations: Group selections where order doesn't matter.
  • Power Set: In our case, the power set is essentially our sample space, each subset representing an event.
Thus, combinatorics is a critical mathematical tool used to evaluate situations like the jury selection, where different arrangements of men and women need to be considered.

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Most popular questions from this chapter

A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let \(x\) equal the number of women chosen to fill the two positions. a. Find \(p(x)\). b. Construct a probability histogram for \(x\).

How many times should a coin be tossed to obtain a probability equal to or greater than .9 of observing at least one head?

If an experiment is conducted, one and only one of three mutually exclusive events \(S_{1}, S_{2},\) and \(S_{3}\) can occur, with these probabilities: $$P\left(S_{1}\right)=.2 \quad P\left(S_{2}\right)=.5 \quad P\left(S_{3}\right)=.3$$ The probabilities of a fourth event \(A\) occurring, given that event \(S_{1}, S_{2},\) or \(S_{3}\) occurs, are $$P\left(A \mid S_{1}\right)=.2 \quad P\left(A \mid S_{2}\right)=.1 \quad P\left(A \mid S_{3}\right)=.3$$ If event \(A\) is observed, find \(P\left(S_{1} \mid A\right), P\left(S_{2} \mid A\right),\) and \(P\left(S_{3} \mid A\right)\).

A man takes either a bus or the subway to work with probabilities .3 and \(.7,\) respectively. When he takes the bus, he is late \(30 \%\) of the days. When he takes the subway, he is late \(20 \%\) of the days. If the man is late for work on a particular day, what is the probability that he took the bus?

A shipping container contains seven complex electronic systems. Unknown to the purchaser, three are defective. Two of the seven are selected for thorough testing and are then classified as defective or nondefective. What is the probability that no defectives are found?
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