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Chapter 2: Problem 82

Suppose that \(A \subset B\) and that \(P(A)>0\) and \(P(B)>0 .\) Show that \(P(B | A)=1\) and \(P(A | B)=P(A) / P(B)\).

Short Answer

Expert verified
\( P(B | A) = 1 \) and \( P(A | B) = \frac{P(A)}{P(B)} \).

Step by step solution

01

Understanding Conditional Probability

In probability theory, the conditional probability of event A given event B is defined as: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]This formula gives the probability of A occurring under the condition that B has occurred.
02

Recognizing Subset Relationship

Since it is given that \( A \subset B \), this tells us that every outcome in A is also in B. Therefore, the intersection \( A \cap B \) is just A itself. So, we have \( P(A \cap B) = P(A) \).
03

Calculate \( P(B | A) \)

Since \( A \subset B \), the occurrence of A guarantees B has occurred. Thus, the probability of B given A is:\[ P(B | A) = \frac{P(B \cap A)}{P(A)} = \frac{P(A)}{P(A)} = 1 \]
04

Calculate \( P(A | B) \)

Using the formula for conditional probability and the results from above, we have:\[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A)}{P(B)} \] since \( P(A \cap B) = P(A) \) from the subset relationship.

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Key Concepts

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

Subset Relationship
In probability theory, understanding the concept of a subset is crucial for analyzing relationships between events. Suppose we have two events, A and B. If A is a subset of B, denoted as \(A \subset B\), it means that every outcome in A is also part of B.
This idea is particularly useful as it simplifies calculations involving these events. When you know one event is a subset of another, it effectively means that if A occurs, B must occur as well.
This subset relationship is vital in calculating conditional probabilities, as it allows us to make confident assertions about the occurrence of events in relation to one another.
Intersection of Events
The intersection of events, represented mathematically as \(A \cap B\), is the set of outcomes that are in both A and B simultaneously. In practical terms, it is the scenario where both events occur together.
In our original problem, since \(A \subset B\), the intersection \(A \cap B\) is just A itself. This is because all elements of A are inherently part of B. Hence, \(P(A \cap B) = P(A)\).
Understanding intersections helps clarify how probabilities are calculated for overlapping events, providing a logical basis for deeper probability computations.
Probability Formula
Conditional probability is an important concept in statistics, allowing us to update probabilities based on new information. The central formula for conditional probability is:
  • \(P(A | B) = \frac{P(A \cap B)}{P(B)}\)
  • This formula gives the likelihood of A occurring given that B has already occurred.

When applying it to our exercise:
  • Since \(A \subset B\), \(P(A \cap B)\) simplifies to \(P(A)\).
  • Thus, \(P(A | B) = \frac{P(A)}{P(B)}\).
The probability formula allows us to break down complex probability scenarios into simpler, more manageable parts.
Mathematical Proof
To provide a concrete understanding of conditional probability, it's essential to construct a mathematical proof using the given conditions. Let's delve into the exercise:
  • We begin by showing \(P(B | A) = 1\).
  • Since \(A \subset B\), the occurrence of A ensures B's occurrence, hence \(P(B \cap A) = P(A)\).
  • Therefore, \(P(B | A) = \frac{P(B \cap A)}{P(A)} = \frac{P(A)}{P(A)} = 1\).
  • Next, for \(P(A | B)\), the subset leads to \(P(A \cap B) = P(A)\) again.
  • The formula becomes \(P(A | B) = \frac{P(A)}{P(B)}\).
Every step of this proof relies on clustering logical deductions based on subset relationships and fundamental probability principles, ensuring you understand the entire process from hypothesis to conclusion.

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Most popular questions from this chapter

A state auto-inspection station has two inspection teams. Team 1 is lenient and passes all automobiles of a recent vintage; team 2 rejects all autos on a first inspection because their "headlights are not properly adjusted." Four unsuspecting drivers take their autos to the station for inspection on four different days and randomly select one of the two teams. a. If all four cars are new and in excellent condition, what is the probability that three of the four will be rejected? b. What is the probability that all four will pass?

Two events \(A\) and \(B\) are such that \(P(A)=.2, P(B)=.3,\) and \(P(A \cup B)=.4 .\) Find the following: a. \(P(A \cap B)\) b. \(P(\bar{A} \cup \bar{B})\) c. \(P(\bar{A} \cap \bar{B})\) d. \(P(\bar{A} | B)\)

Americans can be quite suspicious, especially when it comes to government conspiracies. On the question of whether the U.S. Air Force has withheld proof of the existence of intelligent life on other planets, the proportions of Americans with varying opinions are given in the table. $$\begin{array}{lc} \hline \text { Opinion } & \text { Proportion } \\ \hline \text { Very likely } & .24 \\ \text { Somewhat likely } & .24 \\ \text { Unlikely } & .40 \\ \text { Other } & .12 \\ \hline \end{array}$$ Suppose that one American is selected and his or her opinion is recorded. a. What are the simple events for this experiment? b. Are the simple events that you gave in part (a) all equally likely? If not, what are the probabilities that should be assigned to each? c. What is the probability that the person selected finds it at least somewhat likely that the Air Force is withholding information about intelligent life on other planets?

In the definition of the independence of two events, you were given three equalities to check: \(P(A | B)=P(A)\) or \(P(B | A)=P(B)\) or \(P(A \cap B)=P(A) P(B)\). If any one of these equalities holds, \(A\) and \(B\) are independent. Show that if any of these equalities hold, the other two also hold.

If \(A\) and \(B\) are equally likely events and we require that the probability of their intersection be at least.98, what is \(P(A) ?\)
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