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Show that for any three events \(A, B\), and \(C\) with \(P(C)>0, P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-\) \(P(A \cap B \mid C)\).

Short Answer

Expert verified
The identity is shown using definitions of conditional probability and the union of events formula.

Step by step solution

01

Understand the Problem

We are given three events, \(A\), \(B\), and \(C\) with a given conditional probability that \(P(C) > 0\). We need to demonstrate the truth of the equation \(P(A \cup B \mid C) = P(A \mid C) + P(B \mid C) - P(A \cap B \mid C)\).
02

Recall Definitions and Properties

Recall that the conditional probability of an event \(X\) given another event \(Y\) is defined as \(P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}\), assuming \(P(Y) > 0\). Also, remember that the probability of the union of two events is given by the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
03

Apply the Formula for Conditional Probabilities

First, let's express \(P(A \cup B \mid C)\) using the definition of conditional probability: \[P(A \cup B \mid C) = \frac{P((A \cup B) \cap C)}{P(C)} = \frac{P((A \cap C) \cup (B \cap C))}{P(C)}\].
04

Use the Union of Events Formula

Using the union probability formula, \(P((A \cap C) \cup (B \cap C)) = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)\).
05

Substitute Back into Conditional Probability Formula

Substitute the union formula into the expression for \(P(A \cup B \mid C)\): \[P(A \cup B \mid C) = \frac{P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)}{P(C)}\].
06

Express Each Term as Conditional Probability

Express each term as a conditional probability: \(P(A \mid C) = \frac{P(A \cap C)}{P(C)}\) and \(P(B \mid C) = \frac{P(B \cap C)}{P(C)}\) and \(P(A \cap B \mid C) = \frac{P(A \cap B \cap C)}{P(C)}\).
07

Simplify the Expression

Substitute these expressions back into our equation from Step 5: \[P(A \cup B \mid C) = P(A \mid C) + P(B \mid C) - P(A \cap B \mid C)\]. Thus, we have shown the required identity.

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Key Concepts

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

Probability of Union
The probability of the union of two events, such as events \(A\) and \(B\), is a key concept in probability theory. It tells us how likely it is for either event \(A\), event \(B\), or both events to occur. The formula for determining the probability of the union of two events is:
\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]
This equation helps avoid counting twice the overlapping part, \(P(A \cap B)\), which represents when both events happen together. In the context of conditional probabilities, this principle still applies but is calculated with respect to a given condition or event, such as \(C\).
Hence, the conditional probability of the union \(P(A \cup B \mid C)\) requires dividing the intersection probabilities by \(P(C)\) to reflect the condition that \(C\) occurred.
Intersection of Events
The intersection of two or more events, such as \(A\) and \(B\), represents the scenario where all the events occur together. In probability theory, this is expressed as \(P(A \cap B)\), which measures the likelihood that both \(A\) and \(B\) happen simultaneously.
For conditional probabilities, the approach is similar. For instance, \(P(A \cap B \mid C)\) considers the probability of both \(A\) and \(B\) occurring, given that event \(C\) has happened. This concept is integral in calculating compound probabilities where multiple conditions are imposed.
  • The calculation involves multiplication if events are independent.
  • If dependent, their relationship changes based on how event \(C\) modifies their likelihood.
Conditional Probability Formula
Conditional probability is a fundamental concept in probability theory, representing the probability of an event occurring given that another event has already occurred. It is succinctly expressed by the formula:
\[P(A \mid B) = \frac{P(A \cap B)}{P(B)}\]
This formula calculates the likelihood of event \(A\) given the presence of \(B\), with the assumption that \(P(B) > 0\). It helps in updating the probability of an event as more information becomes available.
  • This formula is particularly useful when the direct probability of an event is unstable or unknown, but other relatable event probabilities are accessible.
  • In conditional scenarios, use intersecting events \((A \cap B)\) to refine probabilities, presenting a more accurate picture of likelihoods under specified conditions.
Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random phenomena. It allows us to quantify uncertainty, analyze data, and make decisions based on probabilities. Core to this theory are events, which can be simple, like flipping a coin, or complex, like predicting weather patterns.
  • Basic rules include those for union, intersection, and conditional probability.
  • Events are often visualized with tools like Venn diagrams or probability trees for better comprehension.
  • Theory is applicable across domains such as finance, science, engineering, and everyday decision-making.
Understanding probability theory offers insights into the likelihood of events, facilitating better risk management and expectation setting in unpredictable scenarios.

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Most popular questions from this chapter

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\\{\) the first board is green \(\\}\) and \(B=\\{\) the second board is green \(\\}\). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=\) \(P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think that an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

Consider randomly selecting a student at a certain university, and let \(A\) denote the event that the selected individual has a Visa credit card and \(B\) be the analogous event for a MasterCard. Suppose that \(P(A)=.5, P(B)=.4\), and \(P(A \cap B)=.25\). a. Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event \(A \cup B\) ). b. What is the probability that the selected individual has neither type of card? c. Describe, in terms of \(A\) and \(B\), the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.

Two voters, \(\mathrm{Al}\) and Bill, are each choosing between one of three candidates \(-1,2\), and \(3-\) who are running for city council. An experimental outcome specifies both Al's choice and Bill's choice, e.g. the pair \((3,2)\). a. List all elements of \(S\). b. List all outcomes in the event \(A\) that \(\mathrm{Al}\) and Bill make the same choice. c. List all outcomes in the event \(B\) that neither of them vote for candidate 2 .

A certain company sends \(40 \%\) of its overnight mail parcels via express mail service \(E_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (denote the event "late delivery" by \(L\) ). If a record of an overnight mailing is randomly selected from the company's file, what is the probability that the parcel went via \(E_{1}\) and was late?

For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+P\left(A^{\prime} \mid B\right)=1 .\)

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