/*! 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 37 Let \(A\) and \(B\) be events. S... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(A\) and \(B\) be events. Show that \(\mathbf{P}(A \cap B \mid A \cup B) \leq \mathbf{P}(A \cap B \mid A)\). When does equality hold?

Short Answer

Expert verified
The inequality holds due to \( \mathbf{P}(A) \geq \mathbf{P}(A \cup B) \), and equality occurs when \( B \subseteq A \).

Step by step solution

01

Recall the definition of conditional probability

Conditional probability is defined as \( \mathbf{P}(X \mid Y) = \frac{\mathbf{P}(X \cap Y)}{\mathbf{P}(Y)} \). This is applicable only when \( \mathbf{P}(Y) > 0 \). Here, two conditional probabilities are involved, \( \mathbf{P}(A \cap B \mid A \cup B) \) and \( \mathbf{P}(A \cap B \mid A) \).
02

Write expressions for conditional probabilities

For \( \mathbf{P}(A \cap B \mid A \cup B) \), we have: \[ \mathbf{P}(A \cap B \mid A \cup B) = \frac{\mathbf{P}((A \cap B) \cap (A \cup B))}{\mathbf{P}(A \cup B)} \] which simplifies to \( \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(A \cup B)} \). For \( \mathbf{P}(A \cap B \mid A) \), we have: \[ \mathbf{P}(A \cap B \mid A) = \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(A)} \].
03

Compare the two probabilities

We want to show that \( \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(A \cup B)} \leq \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(A)} \). This inequality holds when \( \mathbf{P}(A) \geq \mathbf{P}(A \cup B) \).
04

Analyze when the inequality holds exactly

The inequality becomes an equality when \( \mathbf{P}(A) = \mathbf{P}(A \cup B) \). This is only possible when event \( B \) contributes no additional probability outside of \( A \), in other words, when \( B \subseteq A \).
05

Conclusion on the inequality and equality

Based on the inequality \( \mathbf{P}(A \cap B \mid A \cup B) \leq \mathbf{P}(A \cap B \mid A) \), the equality holds when \( B \subseteq A \), suggesting every part of event \( B \) is also part of event \( A \).

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Key Concepts

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

Probability Inequalities
In probability theory, inequalities play a crucial role in comparing the likelihood of different events. When we talk about conditional probabilities, the focus is often on comparing probabilities given certain conditions or events. A fundamental inequality is seen in the relationship between \( \mathbf{P}(A \cap B \mid A \cup B) \) and \( \mathbf{P}(A \cap B \mid A) \). Here, we want to determine if one probability is always less than or equal to the other under certain conditions.

To break it down, consider that \( \mathbf{P}(X \mid Y) \) denotes the probability of event \( X \) occurring given that event \( Y \) has occurred. Using the inequality \( \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(A \cup B)} \leq \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(A)} \) helps demonstrate how probability is affected by the occurrence of different events.
  • This inequality shows that as more information is available (i.e., the event described by \( A \cup B \)), the probability of \( A \cap B \) tends to decrease or remain the same compared to when less information is available (i.e., just the event \( A \)).
  • Such inequalities help us understand how the inclusion of additional conditions or events impacts the likelihood of other events.
Understanding such concepts helps solidify our grasp of how probabilities relate to one another, particularly when assessed under varying conditions.
Set Theory in Probability
The concept of sets forms the foundation of probability theory and is essential for understanding how different events interact. In set theory, events are often represented as sets and operations like intersections (\( \cap \)) and unions (\( \cup \)) depict combinations of these events. Let's explore these concepts in the context of our exercise.

In this problem, we see that \( A \cap B \) and \( A \cup B \) are key expressions.
  • \( A \cap B \) represents the event that both \( A \) and \( B \) happen simultaneously. In terms of sets, this is the intersection of sets \( A \) and \( B \).
  • \( A \cup B \) represents the event that either \( A \) or \( B \) or both occur, which is the union of the sets \( A \) and \( B \).
Set theory in probability helps us visualize and compute probabilities of complex events by relying on these foundational operations.

Furthermore, recognizing the relationship between these sets aids in simplifying expressions and comparing probabilities like in our inequality:
  • The simplification \( \mathbf{P}((A \cap B) \cap (A \cup B)) = \mathbf{P}(A \cap B) \) relies on the property that \( A \cap B \) is naturally within \( A \cup B \).
  • This showcases how set operations not only define the event relationships but also streamline probability calculations.
Through the comprehension of set theory principles, students can gain a clearer understanding of how events overlap and interact within the realm of probability.
Events in Probability
In probability, events essentially depict outcomes or occurrences within a probability space. It's important to realize how these events may influence each other, particularly when looking into their combined or conditional probabilities.

When we focus on the events \( A \) and \( B \), the problem illustrates key aspects of how they relate:
  • An event like \( A \cap B \) indicates simultaneous occurrences, which is crucial for conditional probabilities as it affects the likelihood of occurrences based on joint presence.
  • An event \( A \cup B \) covers more ground by accounting for any outcome within either of the individual events or both, widening the context from which we draw conclusions.
Understanding the scope of events \( A \cap B \) and \( A \cup B \) helps us evaluate probabilities more accurately. This aids in forming effective comparisons critical for conditional expressions.

Looking deeper, we also recognize that both inequality and equality conditions in such probabilities function due to the intrinsic properties of these events:
  • The equality \( \mathbf{P}(A) = \mathbf{P}(A \cup B) \) holds only when the event \( B \) is entirely contained within \( A \), indicating no extra probability contribution from \( B \) beyond \( A \).
  • This clarifies how overlapping and contained events impact probability evaluations, and illustrates the nuanced effects events have on each other.
Effectively recognizing and analyzing these event relationships empowers you to better understand and solve problems involving conditional probabilities.

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Most popular questions from this chapter

Candidates are allowed at most three attempts at a given test. Given \(j-1\) previous failures, the probability that a candidate fails at his \(j\) th attempt is \(p_{j}\). If \(p_{1}=0.6, p_{2}=0.4\), and \(p_{3}=0.75\), find the probability that a candidate: (a) Passes at the second attempt: (b) Passes at the third attempt: (c) Passes given that he failed at the first attempt; (d) Passes at the second attempt given that he passes.

Simpson's Paradox Two drugs are being tested. Of 200 patients given drug \(A, 60\) are cured; and of 1100 given drug \(B, 170\) are cured. If we assume a homogeneous group of patients, find the probabilities of successful treatment with \(A\) or \(B\). Now closer investigation reveals that the 200 patients given drug \(A\) were in fact 100 men, of whom 50 were cured, and 100 women of whom 10 were cured. Further, of the 1100 given drug \(B, 100\) were men of whom 60 were cured, and 1000 were women of whom 110 were cured. Calculate the probability of cure for men and women receiving each drug; note that \(B\) now seems better than \(A\). (Results of this kind indicate how much care is needed in the design of experiments. Note that the paradox was described by Yule in 1903 , and is also called the Yule-Simpson paradox.)

Box \(A\) contains three red balls and two white balls; box \(B\) contains two red balls and two white balls. A fair die is thrown. If the upper face of the die shows 1 or 2 , a ball is drawn at random from box \(A\) and put in box \(B\) and then a ball is drawn at random from box \(B\). If the upper face of the die shows \(3,4,5\) or 6 , a ball is drawn at random from box \(B\) and put in box \(A\), and then a ball is drawn at random from box \(A\). What are the probabilities (a) That the second ball drawn is white? (b) That both balls drawn are red? (c) That the upper face of the red die showed 3, given that one ball drawn is white and the other red?

Weather Days can be sunny or cloudy. The weather tomorrow is the same as the weather today with probability \(p\), or it is different with probability \(q\), where \(p+q=1\). If it is sunny today, show that the probability \(s_{n}\) that it will be sunny \(n\) days from today satisfies $$ s_{n}=(p-q) s_{n-1}+q ; \quad n \geq 1, $$ where \(s_{0}=1\). Deduce that $$ s_{n}=\frac{1}{2}\left(1+(p-q)^{n}\right) ; \quad n \geq 1 $$

Let \(A, B\) be two events with \(\mathbf{P}(B)>0\). Show that (a) If \(B \subset A\), then \(\mathbf{P}(A \mid B)=1\), (b) If \(A \subset B\), then \(\mathbf{P}(A \mid B)=\mathbf{P}(A) / \mathbf{P}(B)\).

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