Probability Of Both Toys Being Defective A Step By Step Solution

Jul 15, 2025 by qnaunigon 65 views

Probability of Defective Toys A Comprehensive Guide

In the realm of probability, understanding conditional probabilities is crucial for solving various real-world problems. This article will delve into a specific problem involving conditional probability, providing a step-by-step solution and exploring the underlying concepts. We'll focus on a scenario where Mrs. Jones purchases two toys for her son, and we need to determine the probability that both toys are defective. Mastering these concepts is not just about acing exams; it's about developing a crucial life skill – the ability to think critically and make informed decisions in the face of uncertainty. This is particularly important in fields like finance, healthcare, and engineering, where risk assessment and probability calculations play a significant role. In finance, for example, understanding the probability of a stock market crash is essential for making sound investment decisions. In healthcare, doctors use probability to assess the likelihood of a patient developing a disease based on their symptoms and risk factors. In engineering, probabilities are used to ensure the safety and reliability of structures and systems. By understanding the fundamentals of probability, you can gain a competitive edge in these fields and make a real-world impact. Furthermore, the concepts discussed here can be applied to everyday situations, such as understanding the odds in a game of chance or making predictions about future events. By honing your probability skills, you can become a more informed and discerning decision-maker, both in your personal and professional life. The problem presented here serves as a microcosm of larger, more complex scenarios, making it an excellent starting point for exploring the fascinating world of probability. So, let's dive in and unravel the intricacies of conditional probability, empowering you with the knowledge and skills to tackle similar problems with confidence.

Problem Statement

Mrs. Jones buys two toys for her son. The probability that the first toy is defective is $\frac{1}{3}$ , and the probability that the second toy is defective given that the first toy is defective is $\frac{1}{8}$ . What is the probability that both toys are defective?

Understanding the Problem

Before diving into the solution, it's crucial to grasp the core concepts. This problem deals with conditional probability, which is the probability of an event occurring given that another event has already occurred. In this case, we're interested in the probability of the second toy being defective given that the first toy is defective. Conditional probability is a fundamental concept in probability theory and statistics, and it plays a crucial role in various fields, including risk assessment, decision-making, and data analysis. It allows us to refine our predictions and make more informed judgments by taking into account prior information. For instance, in medical diagnosis, a doctor might assess the probability of a patient having a certain disease given their symptoms and test results. In finance, investors use conditional probabilities to evaluate the risk of investments based on market conditions and economic indicators. Understanding conditional probability is also essential for interpreting statistical data and drawing meaningful conclusions from research studies. For example, when analyzing the results of a clinical trial, researchers use conditional probabilities to determine the effectiveness of a new treatment compared to a placebo. By mastering conditional probability, you can enhance your ability to analyze complex situations, make informed decisions, and interpret data accurately. This skill is invaluable not only in academic settings but also in professional and personal contexts. So, let's break down the problem step by step and see how conditional probability helps us find the solution.

The problem provides us with two key pieces of information:

The probability of the first toy being defective: $\frac{1}{3}$
The probability of the second toy being defective given that the first toy is defective: $\frac{1}{8}$

Our goal is to find the probability that both toys are defective. This involves understanding how these individual probabilities combine to give us the overall probability of the joint event (both toys being defective). This problem highlights the importance of recognizing the relationship between events. Are they independent, meaning the outcome of one doesn't affect the other? Or are they dependent, as in this case, where the defectiveness of the first toy influences the probability of the second toy being defective? Understanding this distinction is crucial for applying the correct probability rules and arriving at the accurate solution. Furthermore, this problem demonstrates how real-world scenarios can be modeled using probability theory. We're taking a seemingly simple situation – buying toys – and using mathematical tools to analyze the likelihood of a specific outcome. This ability to translate real-world problems into mathematical models is a powerful skill that can be applied to a wide range of situations, from predicting customer behavior to optimizing business processes. By working through this problem, you'll not only learn how to calculate conditional probabilities but also gain a deeper appreciation for the power of probability theory as a problem-solving tool.

Solution

To find the probability that both toys are defective, we use the formula for conditional probability:

$P(A \text{ and } B) = P(A) \times P(B|A)$

Where:

$P(A \text{ and } B)$ is the probability of both events A and B occurring.
$P(A)$ is the probability of event A occurring.
$P(B|A)$ is the probability of event B occurring given that event A has already occurred.

In our case:

Event A: The first toy is defective.
Event B: The second toy is defective.

We are given:

$P(A) = \frac{1}{3}$
$P(B|A) = \frac{1}{8}$

Plugging these values into the formula, we get:

$P(\text{Both toys defective}) = \frac{1}{3} \times \frac{1}{8} = \frac{1}{24}$

Therefore, the probability that both toys are defective is $\frac{1}{24}$ .

Step-by-Step Explanation

Let's break down the solution step-by-step to ensure clarity and understanding. The first step is recognizing that we're dealing with a conditional probability problem. This means the probability of the second toy being defective is dependent on the first toy being defective. This dependence is explicitly stated in the problem: "the probability that the second toy is defective given that the first toy is defective is $\frac{1}{8}$ ." Ignoring this conditional relationship would lead to an incorrect solution. Understanding the nature of the events – whether they are independent or dependent – is crucial for choosing the correct probability formula. In independent events, the outcome of one event does not affect the outcome of the other. For example, flipping a coin twice are independent events. However, in this case, the defectiveness of the first toy influences the probability of the second toy being defective, making them dependent events. Recognizing this distinction is a fundamental aspect of probability theory and is essential for accurate problem-solving.

The second step is recalling the formula for conditional probability: $P(A \text{ and } B) = P(A) \times P(B|A)$ . This formula is the cornerstone of solving this type of problem. It states that the probability of two events A and B both occurring is equal to the probability of event A occurring multiplied by the probability of event B occurring given that event A has already occurred. This formula is derived from the basic principles of probability and is a powerful tool for analyzing situations where events are dependent. It allows us to break down a complex problem into simpler components and calculate the overall probability by considering the individual probabilities and their relationships. This formula is not just applicable to toy defect problems; it can be used in a wide range of scenarios, from predicting the likelihood of a customer making a purchase to assessing the risk of a financial investment. By mastering this formula, you can significantly enhance your problem-solving abilities in various fields.

The third step is identifying the events A and B in the context of the problem. We define event A as "the first toy is defective" and event B as "the second toy is defective." This step is crucial for correctly applying the formula. We need to clearly define what each event represents so that we can substitute the correct probabilities into the equation. This process of translating the problem statement into mathematical terms is a key skill in problem-solving. It involves extracting the relevant information and expressing it in a way that allows us to apply mathematical tools and techniques. In this case, we're transforming the verbal description of the events into concise and unambiguous definitions that can be used in the probability formula. This ability to translate real-world scenarios into mathematical models is a valuable skill that can be applied to various fields, from engineering to finance.

The fourth step is substituting the given probabilities into the formula. We are given that $P(A) = \frac{1}{3}$ (the probability that the first toy is defective) and $P(B|A) = \frac{1}{8}$ (the probability that the second toy is defective given that the first toy is defective). We simply plug these values into the formula: $P(\text{Both toys defective}) = \frac{1}{3} \times \frac{1}{8}$ . This step is a straightforward application of the formula, but it's essential to ensure that the values are substituted correctly. A common mistake is to mix up the probabilities or to not recognize the conditional nature of the probability $\frac{1}{8}$ . By carefully substituting the values, we ensure that we're using the correct information in our calculation and moving closer to the accurate solution. This meticulous approach is crucial for avoiding errors and building confidence in your problem-solving abilities.

The final step is performing the multiplication to obtain the final probability. Multiplying $\frac{1}{3}$ by $\frac{1}{8}$ gives us $\frac{1}{24}$ . This is the probability that both toys are defective. This final calculation is a simple arithmetic operation, but it represents the culmination of all the previous steps. It's the moment where we arrive at the answer to the problem. The probability $\frac{1}{24}$ tells us that, on average, in every 24 instances of Mrs. Jones buying two toys, we would expect both toys to be defective only once. This result highlights the importance of understanding probabilities in real-world situations. It allows us to quantify the likelihood of events and make informed decisions based on those probabilities. By understanding the step-by-step solution, you can not only solve this specific problem but also gain a deeper understanding of conditional probability and its applications.

Key Concepts

This problem illustrates the importance of understanding conditional probability, which is the probability of an event occurring given that another event has already occurred. The formula for conditional probability is a fundamental tool in probability theory and has wide-ranging applications. Mastering this concept is essential for anyone working with data, making decisions under uncertainty, or simply trying to understand the world around them. Conditional probability is used extensively in fields such as statistics, finance, medicine, and engineering. For example, in medical diagnosis, doctors use conditional probabilities to assess the likelihood of a patient having a disease given their symptoms and test results. In finance, investors use conditional probabilities to evaluate the risk of investments based on market conditions and economic indicators. In engineering, conditional probabilities are used to ensure the safety and reliability of systems and structures. Understanding conditional probability also helps us to interpret statistical data and research findings more accurately. For example, when reading about the effectiveness of a new drug, it's important to consider the conditional probability of a patient experiencing side effects given that they are taking the drug. By understanding this concept, we can make more informed decisions and avoid common pitfalls in statistical reasoning.

Another key concept is the distinction between independent and dependent events. Independent events are events where the outcome of one does not affect the outcome of the other. Dependent events, on the other hand, are events where the outcome of one event does influence the outcome of another. In this problem, the events are dependent because the defectiveness of the first toy influences the probability of the second toy being defective. This distinction is crucial for choosing the correct probability formula and avoiding errors in calculations. For example, flipping a coin twice are independent events because the outcome of the first flip does not affect the outcome of the second flip. However, drawing two cards from a deck without replacement are dependent events because the removal of the first card changes the composition of the deck and affects the probability of drawing a specific card on the second draw. Understanding the difference between independent and dependent events is fundamental to probability theory and is essential for accurate problem-solving. It allows us to apply the correct formulas and techniques to calculate probabilities in various scenarios.

Practice Problems

To solidify your understanding, try these practice problems:

A bag contains 5 red balls and 3 blue balls. You draw two balls without replacement. What is the probability that both balls are red?
The probability that it will rain tomorrow is 0.4. The probability that the temperature will be above 80 degrees given that it rains is 0.2. What is the probability that it will rain and the temperature will be above 80 degrees?

Solving these problems will reinforce your understanding of conditional probability and help you apply the concepts learned in this article. Practice is essential for mastering any skill, and probability is no exception. By working through different problems, you'll develop your problem-solving abilities and gain confidence in your understanding of the subject. When solving these problems, remember to carefully identify the events, determine whether they are independent or dependent, and apply the appropriate probability formulas. Don't be afraid to break down the problems into smaller steps and to draw diagrams or use other visual aids to help you visualize the situation. And most importantly, don't give up if you encounter difficulties. Keep practicing, and you'll eventually master the concepts of conditional probability and become a proficient problem-solver.

Conclusion

This article has provided a comprehensive guide to solving a conditional probability problem. By understanding the key concepts and practicing with similar problems, you can confidently tackle probability questions in various contexts. Probability is a powerful tool for understanding and predicting events in the world around us. By mastering the concepts and techniques discussed in this article, you'll be well-equipped to make informed decisions and analyze data effectively. The ability to calculate probabilities and assess risks is a valuable skill in many fields, from finance to medicine to engineering. So, keep practicing, keep exploring, and continue to expand your knowledge of probability – the possibilities are endless.