/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Among 21 - to 25 -year-olds, \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Among 21 - to 25 -year-olds, \(29 \%\) say they have driven while under the influence of alcohol. Suppose three \(21-\) to 25 -year-olds are selected at random. Source: U.S. Department of Health and Human Services, reported in USA Today (a) What is the probability that all three have driven while under the influence of alcohol? (b) What is the probability that at least one has not driven while under the influence of alcohol? (c) What is the probability that none of the three has driven while under the influence of alcohol? (d) What is the probability that at least one has driven while under the influence of alcohol?

Short Answer

Expert verified
(a) 0.0244 (b) 0.9756 (c) 0.3579 (d) 0.6421

Step by step solution

01

- Calculate Probability of a Single Event

The probability that a randomly selected 21- to 25-year-old has driven while under the influence of alcohol is given as 29%, or 0.29. The probability that a person has not driven under the influence is 1 - 0.29 = 0.71.
02

- Calculate Probability of All Three Driving Under Influence

To find the probability that all three have driven while under the influence, use the multiplication rule for independent events. \[ P(\text{all three}) = 0.29 \times 0.29 \times 0.29 = 0.029 \times 0.29 = 0.024389 \]
03

- Calculate Probability of At Least One Not Driving Under Influence

The event 'At least one has not driven while under the influence' is the complement of the event 'all three have driven while under the influence'. Therefore, \[ P(\text{at least one has not driven}) = 1 - P(\text{all three have driven}) = 1 - 0.024389 = 0.975611 \]
04

- Calculate Probability of None Driving Under Influence

To find the probability that none of the three has driven while under the influence, use the multiplication rule for independent events with the complementary probability: \[ P(\text{none have driven}) = 0.71 \times 0.71 \times 0.71 = 0.357911 \]
05

- Calculate Probability of At Least One Driving Under Influence

The event 'At least one has driven while under the influence' is the complement of the event 'none of the three have driven while under the influence'. Therefore, \[ P(\text{at least one has driven}) = 1 - P(\text{none have driven}) = 1 - 0.357911 = 0.642089 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Complement Rule
The complement rule is a foundational concept in probability. To understand it, remember that the total probability of all possible outcomes must equal 1. This means if you know the probability of an event happening, you can easily find the probability of it not happening.
For an event A, the probability of the complement of A (A not happening) is given by: \(P(A^c) = 1 - P(A)\)
In the exercise, the complement rule was used for questions (b) and (d). For instance, question (b) required the probability that at least one has not driven while under the influence. This is the complement of the probability that all three have driven while under the influence. So, we used: \(P(\text{at least one has not driven}) = 1 - P(\text{all three have driven})\)
Similarly, in question (d), to find the probability that at least one has driven while under the influence, we used the complement of the probability that none of them have driven: \(P(\text{at least one has driven}) = 1 - P(\text{none have driven})\). This concept simplifies complex problems by utilizing the relationship between an event and its complement.
Independent Events
Events are independent if the occurrence of one event does not affect the probability of another. Understanding this concept is crucial when dealing with probabilities over multiple trials.
In the given exercise, the probabilities each 21- to 25-year-old has driven under the influence are assumed independent of one another. This means the chance one person has driven under the influence doesn’t change whether another person has or hasn’t.
For probability calculations involving multiple independent events, we multiply the probabilities of each individual event. This principle is applied in questions (a) and (c). In question (a), finding the probability that all three have driven while under the influence requires multiplying the probability for a single event, 0.29, three times: \(P(\text{all three}) = 0.29 \times 0.29 \times 0.29\). For question (c), the same idea is used with the probability of not having driven under the influence: \(P(\text{none have driven}) = 0.71 \times 0.71 \times 0.71\).
This simplicity and relevance make the concept of independent events a cornerstone of probability theory.
Multiplication Rule
The multiplication rule helps in finding the probability that multiple independent events will occur together. As an extension of the concept of independent events, the multiplication rule states: \(P(A \text{ and } B) = P(A) \times P(B)\), provided A and B are independent.
In the exercise, this rule was vital in several calculations. For instance, to find the probability that all three individuals have driven while under the influence, we utilized: \(P(\text{all three}) = 0.29 \times 0.29 \times 0.29\), generating a final result of 0.024389. This application underlines how the multiplication rule magnifies the importance of independence between events.
Every student should understand that this rule only applies when events are independent. Misapplication can lead to errors, especially in real-world scenarios where dependencies might exist.
Event Probability
Event probability is the likelihood or chance that a specific outcome will occur. It ranges between 0 (impossible event) and 1 (certain event). The sum of probabilities of all possible outcomes of a random experiment always equals 1.
In the given problem, the probability of a single event (a person driving under the influence) is 0.29. The event probabilities are fundamental to calculating more complex probabilities involving multiple events.
When multiple individuals or trials are involved, as in questions (a) through (d), understanding how individual event probabilities aggregate (such as through multiplication for independent events) allows for solving larger-scale problems effectively. For instance, knowing \(P(\text{none have driven}) = 0.71 \times 0.71 \times 0.71\) or the previously discussed uses of the complement rule highlights how individual probabilities cohesively build up or influence cumulative event probabilities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose a local area network requires eight characters for a password. The first character must be a letter, but the remaining seven characters can be either a letter or a digit (0 through 9). Lower- and uppercase letters are considered the same. How many passwords are possible for the local area network?

A bag of 100 tulip bulbs purchased from a nursery contains 40 red tulip bulbs, 35 yellow tulip bulbs, and 25 purple tulip bulbs. (a) What is the probability that a randomly selected tulip bulb is red? (b) What is the probability that a randomly selected tulip bulb is purple?

Due to a manufacturing error, three cans of regular soda were accidentally filled with diet soda and placed into a 12-pack. Suppose that two cans are randomly selected from the case. (a) Determine the probability that both contain diet soda. (b) Determine the probability that both contain regular soda. Would this be unusual? (c) Determine the probability that exactly one is diet and one is regular.

Suppose you are dealt 5 cards from a standard 52 -card deck. Determine the probability of being dealt three of a kind (such as three aces or three kings) by answering the following questions: (a) How many ways can 5 cards be selected from a 52 card deck? (b) Each deck contains 4 twos, 4 threes, and so on. How many ways can three of the same card be selected from the deck? (c) The remaining 2 cards must be different from the 3 chosen and different from each other. For example, if we drew three kings, the 4 th card cannot be a king. After selecting the three of a kind, there are 12 different ranks of card remaining in the deck that can be chosen. If we have three kings, then we can choose twos, threes, and so on. Of the 12 ranks remaining, we choose 2 of them and then select one of the 4 cards in each of the two chosen ranks. How many ways can we select the remaining 2 cards? (d) Use the General Multiplication Rule to compute the probability of obtaining three of a kind. That is, what is the probability of selecting three of a kind and two cards that are not like?

Suppose 40 cars start at the Indianapolis \(500 .\) In I how many ways can the top 3 cars finish the race?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.