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Brain Teasers Assume \(A\) and \(B\) are events such that \(0

Short Answer

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False; mutually exclusive events cannot be independent because their joint probability is zero.

Step by step solution

01

Understand the Terms

First, let's understand the terms involved. **Mutually exclusive** means that events A and B cannot happen at the same time, i.e., \( P(A \cap B) = 0 \). **Independent** means that the occurrence of one event does not affect the probability of the other, i.e., \( P(A \cap B) = P(A) \cdot P(B) \).
02

Apply the Definitions

If A and B are mutually exclusive, \( P(A \cap B) = 0 \). For A and B to be independent, \( P(A \cap B) \) must equal \( P(A) \cdot P(B) \). However, since \( P(A \cap B) = 0 \) for mutually exclusive events, it would imply \( P(A) \cdot P(B) = 0 \).
03

Analyze the Probability Condition

Given that \( 0 < P(A) < 1 \) and \( 0 < P(B) < 1 \), multiplying them will not result in zero. Therefore, \( P(A) \cdot P(B) eq 0 \).
04

Conclusion

Since \( P(A \cap B) = 0 \) for mutually exclusive events, and \( P(A \cdot P(B) eq 0 \), mutually exclusive events cannot be independent. The statement is **false**.

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Key Concepts

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

Mutually Exclusive Events
Mutually exclusive events in probability are events that cannot happen at the same time. Think of them like having two doors: if you walk through door A, it's impossible to go through door B at the same moment.This concept is crucial in understanding probability because it directly impacts how we calculate probabilities with such events. When two events are mutually exclusive, their joint probability, represented as \( P(A \cap B) \), is zero. This is because there is no scenario where both events occur simultaneously.
  • Example: Consider tossing a coin. Let event A be getting a "head" and event B be getting a "tail". A coin can't land as both head and tail in one toss, so these events are mutually exclusive.
To calculate the probability of one event or the other occurring (but not both), you simply add their individual probabilities: \( P(A \cup B) = P(A) + P(B) \). This formula is unique to mutually exclusive events because overlapping occurrences do not need to be subtracted.
Independent Events
Independent events are quite different from mutually exclusive ones. Two events are independent if the occurrence of one does not affect the probability of the other event happening. Such events operate independently from each other.
  • Example: Rolling a die and flipping a coin. The result of the die doesn't influence what the coin will land on, and vice versa.
For independent events, the probability of both events occurring is the product of their probabilities, expressed as \( P(A \cap B) = P(A) \cdot P(B) \). This fundamental concept in probability helps in understanding scenarios where multiple trials are independent of each other.Remember, unlike mutually exclusive events, independent events can occur at the same time.
  • Example: Suppose you have \( P(A) = 0.5 \) and \( P(B) = 0.3 \), then \( P(A \cap B) = 0.5 \times 0.3 = 0.15 \).
Basic Probability Rules
Understanding the basic rules of probability is essential to tackling complex problems. These rules serve as the foundation for calculating probabilities in different scenarios. One basic probability rule states that the probability of any event is between 0 and 1. This means that an impossible event has a probability of 0, while a certain event has a probability of 1. Additionally, the sum of probabilities of all possible outcomes of a probability experiment is always 1.
  • Example: In a simple die roll, the sum of probabilities of landing on each face (1-6) is 1.
Moreover, for any two events A and B, the probability of their union is given by the formula: \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). This formula helps account for any overlapping probabilities if events are not mutually exclusive. Understanding these core rules allows one to effectively solve more intricate probability questions by recognizing the interrelations between different events.

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Most popular questions from this chapter

In this problem, you are asked to solve part of the Focus Problem at the beginning of this chapter. In his book Chances: Risk and Odds in Everyday Life, James Burke says that there is a \(72 \%\) chance a polygraph test (lie detector test) will catch a person who is, in fact, lying. Furthermore, there is approximately a \(7 \%\) chance that the polygraph will falsely accuse someone of lying. (a) Suppose a person answers \(90 \%\) of a long battery of questions truthfully. What percentage of the answers will the polygraph wrongly indicate are lies? (b) Suppose a person answers \(10 \%\) of a long battery of questions with lies. What percentage of the answers will the polygraph correctly indicate are lies? (c) Repeat parts (a) and (b) if \(50 \%\) of the questions are answered truthfully and \(50 \%\) are answered with lies. (d) Repeat parts (a) and (b) if \(15 \%\) of the questions are answered truthfully and the rest are answered with lies.

Brain Teasers Assume \(A\) and \(B\) are events such that \(0

Brain Teasers Assume \(A\) and \(B\) are events such that \(0

Greg made up another question for a small quiz. He assigns the probabilities \(P(A)=0.6, P(B)=0.7, P(A | B)=0.1\) and asks for the probability \(P(A \text { or } B\) ). What is wrong with the probability assignments?

Marketing: Toys USA Today gave the information shown in the table about ages of children receiving toys. The percentages represent all toys sold. What is the probability that a toy is purchased for someone (a) 6 years old or older? (b) 12 years old or younger? (c) between 6 and 12 years old? (d) between 3 and 9 years old? $$\begin{array}{l|c} \hline \text { Age (years) } & \text { Percentage of Toys } \\ \hline 2 \text { and under } & 15 \% \\ 3-5 & 22 \% \\ 6-9 & 27 \% \\ 10-12 & 14 \% \\ 13 \text { and over } & 22 \% \\ \hline \end{array}$$ A child between 10 and 12 years old looks at this probability distribution and asks, "Why are people more likely to buy toys for kids older than I am \([13\) and over] than for kids in my age group \([10-12] ?^{\prime \prime}\) How would you respond?

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