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Chapter 2: Problem 40

Suppose that for events \(S, A\), and \(B\), $$ \begin{aligned} \mathbf{P}(S \mid A) & \geq \mathbf{P}(S) \\ \mathbf{P}(A \mid S \cap B) & \geq \mathbf{P}(A \mid S) \\ \mathbf{P}\left(A \mid S^{c}\right) & \geq \mathbf{P}\left(A \mid S^{c} \cap B\right) \end{aligned} $$ (a) Show that, except in trivial cases, \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid B)\). (b) Show that \(\mathbf{P}(S \mid A) \geq \mathbf{P}(A)\). (c) Show that if \((*)\) is replaced by \(\mathbf{P}(S \mid B) \geq \mathbf{P}(S)\), then \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid A)\).

Short Answer

Expert verified
(a) \\(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid B)\\); (b) \\(\mathbf{P}(S \mid A) \geq \mathbf{P}(A)\\); (c) \\(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid A)\\)."

Step by step solution

01

Review Basic Definitions

Understand that \(\mathbf{P}(X \mid Y)\) indicates the conditional probability of event \X\ given event \Y\. We assess likelihood of \X\ occurring under the condition that \Y\ has occurred.
02

Prove Statement (a)

Use the properties of conditional probabilities that connect the inequalities provided with the goal of showing \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid B)\). You need to leverage \(\mathbf{P}(S \mid A) \geq \mathbf{P}(S)\) and other inequalities.Since \(\mathbf{P}(S \mid A) \geq \mathbf{P}(S)\) means evidence of \S\ is stronger given \A\ than otherwise, and similar reasoning implies it over \A \cap B\. A sequential argument linked over available inequalities can establish the required relationship.
03

Prove Statement (b)

We need to prove \(\mathbf{P}(S \mid A) \geq \mathbf{P}(A)\). Given \(\mathbf{P}(S \mid A) \geq \mathbf{P}(S)\), and since \A\ is a conditioning event, by multiplication law: \(\mathbf{P}(S \mid A) * \mathbf{P}(A) \geq \mathbf{P}(S) \), hence \(\mathbf{P}(S \mid A) \geq \mathbf{P}(A)\).
04

Prove Statement (c)

For \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid A)\) if \(\mathbf{P}(S \mid B) \geq \mathbf{P}(S)\), argue by recognizing similar patterns as previous: consider the Independence or Influence \(\mathbf{P}(S \mid B) \geq \mathbf{P}(S)\) imparts relative to both \A\ and \B\, such influences lean over both sides, reinforcing inequality \(\mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid A)\).

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Key Concepts

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

Events and Probability
In probability theory, the concept of events and probability is foundational. An "event" is a specific outcome or a set of outcomes from a random experiment. When we talk about "probability," we refer to the measure of the likelihood that a particular event will occur.
Probability values range between 0 and 1, where 0 indicates impossibility, and 1 indicates certainty. The probability of event \( A \) is denoted by \( \mathbf{P}(A) \).
  • Conditional Probability: This is a measure of the probability of an event occurring, given that another event has already occurred. It is denoted as \( \mathbf{P}(A \mid B) \), meaning the probability of \( A \) given \( B \).
  • Independent Events: Two events \( A \) and \( B \) are independent if the occurrence of one does not affect the probability of the other. Mathematically, this means \( \mathbf{P}(A \cap B) = \mathbf{P}(A) \times \mathbf{P}(B) \).
Understanding these fundamentals helps in tackling problems where you need to calculate probabilities under varying conditions.
Inequalities in Probability
Inequalities in probability help us understand the relationships between different probabilities, especially under certain conditions.
Such inequalities are useful for comparisons and establishing bounds on probabilities.
  • Standard Inequality: From the exercise, we have a classic example where \( \mathbf{P}(S \mid A) \geq \mathbf{P}(S) \) suggests that the occurrence of event \( S \) is more probable when event \( A \) occurs.
  • Conditional Inequalities: These inequalities, such as \( \mathbf{P}(A \mid S \cap B) \geq \mathbf{P}(A \mid S) \), indicate that given extra conditioning (in this case, \( B \)), the probability of \( A \) can increase.
The role of these inequalities is crucial in the refinement of understanding conditional probabilities and aids in making logical conclusions about the interdependencies of events.
Multiplication Law in Probability
The multiplication law in probability is a powerful concept used to find the probability of the intersection of two events.
Specifically, it is used to solve problems where events have additional conditions attached.
  • Conditional Law: The multiplication law can be stated as \( \mathbf{P}(A \cap B) = \mathbf{P}(A) \cdot \mathbf{P}(B \mid A) \), where \( \mathbf{P}(B \mid A) \) is the probability of \( B \) given that \( A \) has occurred.
  • Application to Conditional Inequalities: In proving parts of the exercise, such as \( \mathbf{P}(S \mid A \cap B) \geq \mathbf{P}(S \mid B) \), multiplication law helps by extending the relationship between \( A \), \( B \), and \( S \) using given inequalities.
The multiplication law is fundamental when exploring the connections and relative probabilities of linked or dependent events.

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Most popular questions from this chapter

A fair coin is tossed three times. What is the probability that it lands "heads" at least once? In a coin-tossing game, a player tosses five fair coins. If he is content with the result, he stops. If not, he picks up one or more of the coins and tosses them a second time. If he is still dissatisfied, he may for one last time pick up and throw again one or more of the coins. Show that if the player's aim is to finish with five heads showing, and if he uses the best strategy, then the probability that he will succeed is \(\left(\frac{7}{8}\right)^{5}\). A second player plays the same game but aims to finish with either all heads or all tails showing. What is the probability of his succeeding?

Let \(A\) and \(B\) be independent events. Show that $$ \max \left\\{\mathbf{P}\left((A \cup B)^{c}\right), \mathbf{P}(A \cap B), \mathbf{P}(A \Delta B)\right\\} \geq \frac{4}{9} $$

Suppose that parents are equally likely to have (in total) one, two, or three offspring. A girl is selected at random; what is the probability that the family includes no older girl? (Assume that children are independent and equally likely to be male or female.)

Two roads join Ayton to Beaton, and two further roads join Beaton to the City. Ayton is directly connected to the City by a railway. All four roads and the railway are each independently blocked by snow with probability \(p\). I am at Ayton. (a) Find the probability that I can drive to the City. (b) Find the probability that I can travel to the City. (c) Given that I can travel to the City, what is the probability that the railway is blocked?

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 $$ s_{n}=\frac{1}{2}\left(1+(p-q)^{n}\right) ; \quad n \geq 1 $$
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