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

Short Answer

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True, because events \(A\) and \(A^c\) are mutually exclusive.

Step by step solution

01

Understand Event and Complement

In probability, an event like \(A\) has a complement event, denoted as \(A^c\). The event \(A^c\) represents all outcomes not in \(A\). Therefore, \(A\) and \(A^c\) are mutually exclusive events, meaning they cannot occur simultaneously.
02

Define Probability of Intersection

The probability of the intersection of \(A\) and \(A^c\) is expressed as \(P(A \cap A^c)\). The intersection represents outcomes that are in both \(A\) and \(A^c\). Since \(A\) and \(A^c\) are mutually exclusive, there are no common outcomes.
03

Apply Probability Rule for Mutually Exclusive Events

For mutually exclusive events, the probability of their intersection is zero because they do not share any outcomes. Therefore, \(P(A \cap A^c) = 0\). This implies that the event \(A\) occurring and the event \(A^c\) occurring at the same time is impossible.

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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 are situations where two events cannot happen at the same time. As an example, consider flipping a coin: the result can either be heads or tails, but never both. This is a perfect illustration of mutually exclusive events. In mathematical terms, if two events, say \(A\) and \(B\), are mutually exclusive, they do not overlap. That means the probability of both \(A\) happening and \(B\) happening simultaneously is zero: \[ P(A \cap B) = 0 \]This is because there is no shared outcome between the two events. Understanding this helps solve many probability questions as it simplifies the calculation of probabilities involving these two events.
Event and Complement
In probability, each event has a counter-event known as a complement. This complement event contains all outcomes not included in the main event. Suppose we have an event \(A\). Its complement is represented as \(A^c\). For example, if event \(A\) is drawing a red card from a deck, then \(A^c\) is drawing a card that is not red. These two events, \(A\) and \(A^c\), are mutually exclusive by definition. They cannot occur at the same time. The complement rule is a crucial part of probability calculations:\[ P(A) + P(A^c) = 1 \]This means the probability that event \(A\) will happen, plus the probability that it won’t happen, must equal one. It's a straightforward rule, yet incredibly powerful for understanding complex probabilistic systems.
Probability of Intersection
The probability of an intersection between two events involves calculating the likelihood that both events occur simultaneously. In mathematical notation, this is shown as \(P(A \cap B)\), which denotes the intersection of events \(A\) and \(B\). For non-mutually exclusive events, this calculation includes any outcomes that both events share. However, if the events are mutually exclusive, like an event \(A\) and its complement \(A^c\), the probability of the intersection is zero:\[ P(A \cap A^c) = 0 \]This zero probability occurs because no outcomes are shared between \(A\) and \(A^c\), reinforcing that they cannot occur at the same time. Understanding intersections in probability can help solve complex problems, especially when determining how different possible events interact with each other.

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Most popular questions from this chapter

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?

Environmental: Land Formations Arches National Park is located in southern Utah. The park is famous for its beautiful desert landscape and its many natural sandstone arches. Park Ranger Edward McCarrick started an inventory (not yet complete) of natural arches within the park that have an opening of at least 3 feet. The following table is based on information taken from the book Canyon Country Arches and Bridges by F. A. Barnes. The height of the arch opening is rounded to the nearest foot. $$\begin{array}{l|ccccc} \hline \text { Height of arch, feet } & 3-9 & 10-29 & 30-49 & 50-74 & 75 \text { and higher } \\ \hline \begin{array}{l} \text { Number of arches } \\ \text { in park } \end{array} & 111 & 96 & 30 & 33 & 18 \\ \hline \end{array}$$ For an arch chosen at random in Arches National Park, use the preceding information to estimate the probability that the height of the arch opening is (a) 3 to 9 feet tall (b) 30 feet or taller (c) 3 to 49 feet tall (d) 10 to 74 feet tall (e) 75 feet or taller

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

Based on data from the Statistical Abstract of the United States, 112 th edition, only about \(14 \%\) of senior citizens \((65\) years old or older) get the flu each year. However, about \(24 \%\) of the people under 65 years old get the flu each year. In the general population, there are \(12.5 \%\) senior citizens \((65\) years old or older). (a) What is the probability that a person selected at random from the general population is a senior citizen who will get the flu this year? (b) What is the probability that a person selected at random from the general population is a person under age 65 who will get the flu this year? (c) Answer parts (a) and (b) for a community that is \(95 \%\) senior citizens. (d) Answer parts (a) and (b) for a community that is \(50 \%\) senior citizens.

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?

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