/*! 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 5 Suppose that \(21 \%\) of people... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that \(21 \%\) of people own dogs. If you pick two people at random, what is the probability that neither own a dog?

Short Answer

Expert verified
The probability that neither owns a dog is 0.6241.

Step by step solution

01

Understand the Problem

We need to find the probability that neither of the two people selected at random owns a dog, given that 21% of people own dogs.
02

Calculate Probability of One Person Not Owning a Dog

If 21% of people own dogs, then 79% do not own dogs. This is calculated by subtracting the probability of owning a dog from 1: \[ P( ext{not owning a dog}) = 1 - 0.21 = 0.79 \]
03

Use Independent Probability Concept

Since the two events (each person not owning a dog) are independent, the probability that both events occur is the product of their individual probabilities.
04

Calculate the Probability of Neither Owning a Dog

Multiply the probability that one person does not own a dog by itself to find the probability that neither person owns a dog. \[ P( ext{neither owns a dog}) = 0.79 \times 0.79 = 0.6241 \]

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Key Concepts

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

Independent Events in Probability Theory
In probability theory, understanding whether events are independent is crucial for calculating combined probabilities accurately. Two events are considered independent if the outcome of one event does not influence the outcome of the other.
This means that the probability of both events happening is simply the product of their individual probabilities.
Imagine flipping a coin twice. The outcome of the first flip (heads or tails) doesn't affect the outcome of the second flip. Similarly, in this exercise, whether the first person owns a dog does not influence whether the second person owns a dog. Therefore, the events are independent.
This allows us to multiply their probabilities directly when calculating the chance that neither of two randomly picked individuals own a dog. This fundamental principle makes many real-world probability calculations possible.
Basic Probability Calculation
Probability calculation is a way to quantify the likelihood of an event occurring. Probabilities range from 0 to 1, where 0 means an event is impossible, and 1 means it is certain. The probability of an event can be expressed using the formula:\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]For example, if 21% of people own dogs, we can say the probability of a randomly selected person owning a dog is 0.21.
Consequently, the probability of the complementary event, i.e., not owning a dog, can be found using another basic probability formula:\[ P(\text{not owning a dog}) = 1 - P(\text{owning a dog}) = 1 - 0.21 = 0.79 \]This approach helps us understand and solve many basic probability problems efficiently.
Complementary Probability and Its Importance
Complementary probability is a useful concept when dealing with probabilities that add up to 1. For any event, its complement is the event that the original event does not occur. The sum of their probabilities is always equal to 1.
Consider the situation where 21% of people own dogs. The complement of this event is that a person does not own a dog, which we calculated as 79% since:\[ P(\text{not owning a dog}) = 1 - P(\text{owning a dog}) = 1 - 0.21 = 0.79 \]Complementary probability helps simplify complex probability problems, especially when involved in multiple calculations.
By understanding the complementary probabilities, we can work backwards from what is not happening to deduce what is happening, offering a straightforward way to ensure our calculations cover all possibilities.

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Most popular questions from this chapter

A friend devises a game that is played by rolling a single six-sided fair (each side has equal probability of landing face up, once rolled) die once. If you roll a \(6,\) he pays you \(\$ 3:\) if you roll a \(5,\) he pays you nothing; if you roll a number less than 5 , you pay him \(\$ 1\). a. Make a probability model for this game. b. Compute the expected value for this game. c. Should you play this game?

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A ball is drawn randomly from a jar containing 12 red marbles, 8 white marbles, and 5 yellow marbles. Find the probability of: a. Drawing a red marble. b. Not drawing a white marble. c. Drawing a yellow or red marble. d. Drawing a blue marble. e. Drawing two red marbles if you draw with replacement. f. Drawing first a red marble then a yellow marble if marbles are drawn without replacement.

What is the probability of rolling two fair six-sided dice a. and getting a sum greater than or equal to \(7 ?\) b. getting an even sum or a sum greater than \(7 ?\)

A bag contains 3 gold marbles, 6 silver marbles, and 28 black marbles. Someone offers to play this game: You randomly select on marble from the bag. If it is gold, you win \(\$ 3 .\) If it is silver, you win \(\$ 2 .\) If it is black, you lose \(\$ 1 .\) a. Make a probability model for this game. b. What is your expected value if you play this game?

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