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Use the axioms to show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B\) ), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B), P(A)\), and \(P(A \cup B)\) ?

Short Answer

Expert verified
If \(A \subseteq B\), then \(P(A) \leq P(B)\). For any \(A\) and \(B\), we have \(P(A \cap B) \leq P(A), P(B)\) and \(P(A \cup B) \geq P(A), P(B)\).

Step by step solution

01

Understanding the Problem

We need to prove that if an event \(A\) is a subset of another event \(B\), then the probability of \(A\) is less than or equal to the probability of \(B\). We are given a hint: \(B = A \cup (B \cap A')\), where \(A'\) is the complement of \(A\), and \(A\) and \(B \cap A'\) are disjoint.
02

Apply the Subset Property

Since \(A\) is a subset of \(B\), every outcome in \(A\) is also in \(B\). Thus, \(B\) can be expressed as the union of \(A\) and some other part \((B \cap A')\), which does not intersect with \(A\). This can be seen with a Venn diagram, where \(B\) surrounds \(A\) completely.
03

Use the Additivity of Probability

From the hint, \(B = A \cup (B \cap A')\) and \(A\) and \(B \cap A'\) are disjoint, therefore by the Axiom of Additivity, we have:\[ P(B) = P(A) + P(B \cap A') \]Since \(P(B \cap A')\) is non-negative, it follows that \(P(B) \geq P(A)\).
04

Consider General Relationship Among A, B, A ∩ B and A ∪ B

For general events \(A\) and \(B\), \(A \cap B\) is the intersection of \(A\) and \(B\), and \(A \cup B\) represents all outcomes in either \(A\), \(B\), or both. Therefore, \(P(A \cap B) \leq P(A)\) and \(P(A \cap B) \leq P(B)\) since the intersection is part of both \(A\) and \(B\). Also, \(P(A)\) and \(P(B)\) must be less than or equal to \(P(A \cup B)\) by the additive property of probability, as they contribute to \(P(A \cup B)\).

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Key Concepts

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

Set Theory
In probability theory, set theory serves as a foundational tool to describe phenomena associated with random experiments. An *event* is a set of outcomes occurring from a random experiment. The concept of sets becomes crucial when understanding relationships between different events. Using sets, we define complex operations like unions, intersections, and complements between events.
A set can encompass measurable elements that form the basis for calculating probabilities. For instance, if an event \(A\) is a subset of an event \(B\), every outcome in \(A\) is also in \(B\). This is denoted by \(A \subseteq B\), highlighting the idea that knowledge of set structures is pivotal in analyzing probability relations.
Additive Property
One of the axioms of probability regarded as the additive property is essential for calculating the probability of unions of disjoint events. When two events, \(A\) and \(B\), are disjoint (i.e., \(A \cap B = \emptyset\)), the probability of either event occurring is the sum of their individual probabilities:
  • \(P(A \cup B) = P(A) + P(B)\)
This property forms the basis for extending probabilities for any number of disjoint events.
For instance, if we express \(B\) as \(A \cup (B \cap A')\), where \(A\) and \(B \cap A'\) are disjoint, the additive property assures us that \(P(B) = P(A) + P(B \cap A')\). This relationship affirms that the total probability of \(B\) encompasses the probability of \(A\) and any additional outcomes in \(B\) that are not in \(A\).
Axioms of Probability
Axioms of probability form the foundation from which all probability rules derive. The three key axioms are:
  • Non-negativity: \(P(E) \geq 0\) for any event \(E\).
  • Normalization: \(P(S) = 1\), where \(S\) is the sample space containing all possible outcomes.
  • Additivity: If \(A\) and \(B\) are disjoint events, then \(P(A \cup B) = P(A) + P(B)\).
These axioms allow us to systematically compute probabilities and understand complex relationships between events.
In their application, such as proving \(P(A) \leq P(B)\) when \(A \subseteq B\), they provide means to apply properties like additivity to argue that additional outcomes in \(B\) ensure its probability can only be equal to or greater than that of \(A\).
Venn Diagram
A Venn diagram is a visual tool used in probability and set theory to illustrate the relationships and interactions between different sets or events. Regions within the circles or ovals in a Venn diagram represent events, while the overlap indicates their intersections.
For example, consider events \(A\) and \(B\). Using a Venn diagram, you can visually depict that \(A\) is a subset of \(B\) by placing \(A\) entirely within the boundary of \(B\). This representation helps students easily visualize concepts like unions and intersections.
  • In solving exercises, seeing how \(B = A \cup (B \cap A')\), the Venn diagram confirms the expression by showing \(B\) surrounding \(A\) with any difference accounted for by \(B \cap A'\), an essential tool in understanding such relational proofs.
Event Intersection
Intersections are a critical concept in probability, denoting the probability that both of two events happen simultaneously. The intersection \(A \cap B\) refers to the set of outcomes shared by both events \(A\) and \(B\).
Understanding intersections helps in solving problems related to joint probability and explaining relationships between events. For instance, with any two arbitrary events \(A\) and \(B\), the probability of their intersection \(P(A \cap B)\) is less than or equal to either \(P(A)\) or \(P(B)\). This is because the intersection can't be larger than the individual sets.
  • A fundamental aspect of this concept is its role in applications of conditional probability and set operations in probability theory.
Overall, properly comprehending intersections enriches our understanding of how events relate within a sample space.

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Most popular questions from this chapter

A company uses three different assembly lines- \(A_{1}\), \(A_{2}\), and \(A_{3}\) - to manufacture a particular component. Of those manufactured by line \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components need rework and \(10 \%\) of \(A_{3}\) 's need rework. Suppose that \(50 \%\) of all components are produced by line \(A_{1}\), \(30 \%\) are produced by line \(A_{2}\), and \(20 \%\) come from line \(A_{3}\). If a randomly selected component needs rework, what is the probability that it came from line \(A_{1}\) ? From line \(A_{2}\) ? From line \(A_{3}\) ?

A chemist is interested in determining whether a certain trace impurity is present in a product. An experiment has a probability of \(.80\) of detecting the impurity if it is present. The probability of not detecting the impurity if it is absent is \(.90\). The prior probabilities of the impurity being present and being absent are \(.40\) and \(.60\), respectively. Three separate experiments result in only two detections. What is the posterior probability that the impurity is present?

Show that \(\left(\begin{array}{c}n \\\ k\end{array}\right)=\left(\begin{array}{c}n \\ n-k\end{array}\right)\). Give an interpretation involving subsets.

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost . Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Manag., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

Fasteners used in aircraft manufacturing are slightly crimped so that they lock enough to avoid loosening during vibration. Suppose that \(95 \%\) of all fasteners pass an initial inspection. Of the \(5 \%\) that fail, \(20 \%\) are so seriously defective that they must be scrapped. The remaining fasteners are sent to a recrimping operation, where \(40 \%\) cannot be salvaged and are discarded. The other \(60 \%\) of these fasteners are corrected by the recrimping process and subsequently pass inspection. a. What is the probability that a randomly selected incoming fastener will pass inspection either initially or after recrimping? b. Given that a fastener passed inspection, what is the probability that it passed the initial inspection and did not need recrimping?

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