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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 \(\mathbf{P}(A \mid B)=\mathbf{P}(A) / \mathbf{P}(B)\).

Short Answer

Expert verified
(a) \(\mathbf{P}(A \mid B)=1\); (b) \(\mathbf{P}(A \mid B)=\mathbf{P}(A)/\mathbf{P}(B)\).

Step by step solution

01

Understanding Conditional Probability

The conditional probability \(\mathbf{P}(A \mid B)\) represents the probability of event \(A\) occurring given that event \(B\) has occurred. It is defined as \(\mathbf{P}(A \mid B) = \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(B)}\).
02

Verification for Part (a)

Given that \(B \subset A\), it implies that whenever \(B\) occurs, \(A\) must also occur. Thus, \(A \cap B = B\). Substituting this into the formula for conditional probability gives:\[\mathbf{P}(A \mid B) = \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(B)} = \frac{\mathbf{P}(B)}{\mathbf{P}(B)} = 1.\]
03

Verification for Part (b)

Given that \(A \subset B\), \(A \cap B = A\). Substituting this into the formula for conditional probability gives:\[\mathbf{P}(A \mid B) = \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(B)} = \frac{\mathbf{P}(A)}{\mathbf{P}(B)}.\]

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Key Concepts

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

Events in Probability
In probability theory, an event is essentially a set of outcomes from a probability space. Think of tossing a coin: you could have events such as getting a heads or not getting a heads. Just like events like these, we also deal with more complex events in probability. An event is often denoted by a capital letter such as \(A\) or \(B\). Each of these letters represents a particular scenario or result we're interested in. When solving probability problems, understanding the nature of these events is crucial.

It's important to differentiate between simple events and compound events. Simple events cannot be broken down further, while compound events are composed of two or more simple events. For example, getting a heads on a single coin toss is a simple event. In contrast, getting a heads two times in two coin tosses is a compound event, involving two simple events occurring together.

Events are the building blocks of probability problems and become even more interesting when we start looking into their relationships with each other, which is where subset relations come into play.
Subset Relations
Subset relations in probability relate to how different events are connected to each other. If we say that event \(B\) is a subset of event \(A\) (denoted as \(B \subset A\)), it means that whenever \(B\) happens, \(A\) must also happen. This relationship is pivotal in understanding how the occurrence of one event can influence the probability of another event.

When an event is a subset of another, it hints at a dependent relationship. This is important in solving conditional probability problems. If event \(A\) is known to occur whenever event \(B\) occurs, it simplifies the probability calculations significantly. For example, if \(B \subset A\), then every outcome in \(B\) is naturally included in \(A\), leading to simplified equations. Subset relations are foundational in solving probability exercises as it often helps in reducing the complexity of the calculations.
Probability Formulas
Formulas play a key role in solving probability problems and deriving solutions. One fundamental formula in probability is the conditional probability formula: \[ \mathbf{P}(A \mid B) = \frac{\mathbf{P}(A \cap B)}{\mathbf{P}(B)} \]This formula calculates the probability of event \(A\) occurring given that event \(B\) has already occurred. This is known as the conditional probability and is crucial for solving problems where events are dependent on each other.

In the context of subset relations:
  • If \(B \subset A\), it simplifies further to \( \mathbf{P}(A \mid B) = 1 \) because \( \mathbf{P}(A \cap B) = \mathbf{P}(B) \). Every time \(B\) happens, \(A\) must happen, so the probability is always 1.
  • If \(A \subset B\), it changes to \( \mathbf{P}(A \mid B) = \frac{\mathbf{P}(A)}{\mathbf{P}(B)} \) because \( \mathbf{P}(A \cap B) = \mathbf{P}(A) \).
Understanding these formulas and when to apply them, based on the subset relations, simplifies many complex probability problems. Formulas are not just shortcuts but tools for understanding how different probabilistic scenarios are connected.

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Most popular questions from this chapter

Three biased coins \(C_{1}, C_{2}, C_{3}\) lie on a table. Their respective probabilities of falling heads when tossed are \(\frac{1}{3}, \frac{2}{3}\), and 1 . A coin is picked at random, tossed, and observed to fall heads. Calculate the probability that it is \(C_{k}\) for each \(k=1,2,3\). Given that a coin has been tossed once and observed to fall heads, calculate the probability that a second throw of the same coin will also produce heads. The experiment is begun again with the same three coins. This time the coin selected is tossed twice and observed to fall heads both times. Calculate the probability that it is \(C_{k}\) for each \(k=1,2,3\). Given that a coin has been tossed twice and observed to fall heads both times, calculate the probability that a third throw of the same coin will also produce heads.

A 12 -sided die \(A\) has 9 green faces and 3 white faces, whereas another 12 -sided die \(B\) has 3 green faces and 9 white faces. A fair coin is tossed once. If it falls heads, a series of throws is made with die \(A\) alone; if it falls tails then only the die \(B\) is used. (a) Show that the probability that green turns up at the first throw is \(\frac{1}{2}\). (b) If green turns up at the first throw, what is the probability that die \(A\) is being used? (c) Given that green turns up at the first two throws, what is the probability that green turns up at the third throw?

Suppose that any child is male with probability \(p\) or female with probability \(1-p\), independently of other children. In a family with four children, let \(A\) be the event that there is at most one girl, and \(B\) the event that there are children of both sexes. Show that there is a value of \(p\), with \(0

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.

\(A\) and \(B\) each have $$\$ 60$$. They play a sequence of independent games at each of which \(A\) wins \(\$ x\) from \(B\) with probability \(p\), or loses \(\$ x\) to \(B\) with probability \(q\), where \(p+q=1\). The stake \(x\) is determined by rolling a fair die once, and setting \(x\) as the number shown by the die; \(1 \leq x \leq 6\) (a) What is the probability that \(A\) wins his opponent's fortune before losing his own? (b) If \(A\) could choose the stake to be an integer \(x\) such that \(1 \leq x \leq 6\), and \(p

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