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

Short Answer

Expert verified
The sum \(P(A \mid B) + P(A' \mid B)\) equals 1.

Step by step solution

01

Understand the Definitions

In probability theory, the conditional probability \( P(A \mid B) \) is the probability of event \( A \) occurring given that \( B \) has occurred. Similarly, \( P(A' \mid B) \) is the probability of event \( A' \) occurring given that \( B \) has occurred, where \( A' \) is the complement of \( A \).
02

Use the Formula for Conditional Probability

The formula for conditional probability is \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \) and \( P(A' \mid B) = \frac{P(A' \cap B)}{P(B)} \). Here, \( A \cap B \) represents the intersection of events \( A \) and \( B \), and \( A' \cap B \) represents the intersection of \( A' \) and \( B \).
03

Apply the Complement Rule

Note that \( A \cup A' = \Omega \) (the entire sample space) and \( A \cap A' = \emptyset \) (the empty set). Thus, \( A \cap B \) and \( A' \cap B \) form a partition of \( B \). Therefore, \( P(A \cap B) + P(A' \cap B) = P(B) \).
04

Derive the Equation

Substitute the results from Step 3 into the formulas for conditional probabilities: \[P(A \mid B) + P(A' \mid B) = \frac{P(A \cap B)}{P(B)} + \frac{P(A' \cap B)}{P(B)} = \frac{P(A \cap B) + P(A' \cap B)}{P(B)}\]
05

Simplify the Expression

Using the identity from Step 3, substitute \( P(A \cap B) + P(A' \cap B) = P(B) \) into the equation above to get: \[P(A \mid B) + P(A' \mid B) = \frac{P(B)}{P(B)}\] Simplify this to \( 1 \).

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Key Concepts

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

Complement of an Event
In probability theory, understanding the complement of an event is crucial in determining probabilities and outcomes. The complement of an event, say event \(A\), is another event, denoted as \(A'\), which includes all possible outcomes that are not in \(A\). To simplify, if you know event \(A\) signifies selecting a red ball from a bag, then the complement \(A'\) would represent choosing any ball that is not red.
The concept of complement helps in various probability calculations, especially when determining the chance that either one event occurs or it doesn't, which is always equal to 1. This is because the complete sample space, denoted as \(\Omega\), is equal to the union of \(A\) and \(A'\) with no overlap, represented by \(A \cup A' = \Omega\) and \(A \cap A' = \emptyset\).By understanding and using complements, one can solve probability problems efficiently without having to always calculate every specific outcome.
Probability Theory
Probability theory is a fundamental branch of mathematics that deals with uncertainty and randomness. It helps us calculate the likelihood of different outcomes. Whether it’s flipping a coin, rolling a die, or picking a card from a deck, probability theory is used to determine the chance of specific results.
The entire realm of possibilities in a given situation is known as the sample space \(\Omega\). Each specific event is a subset of this sample space. Events can be independent, dependent, or mutually exclusive. In our context, we’re interested in dependent events, where the occurrence of one affects the probability of another, as seen with conditional probabilities.
Fundamental rules include the additive rule \(P(A \cup B)\), used when considering the union of two events, and the multiplicative rule \(P(A \cap B)\) for their intersection. These concepts provide the foundation for more complex models and applications, used widely in fields ranging from finance to artificial intelligence.
Conditional Probability Formula
Conditional probability is a concept used to find the probability of an event occurring under a certain condition. Expressed as \(P(A \mid B)\), it represents the probability of event \(A\) happening given that event \(B\) has occurred. The formula used is:\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \]This formula calculates how likely \(A\) is when \(B\) has already taken place, using the intersection \(A \cap B\), which includes only outcomes common to both events, divided by the probability of \(B\).
The conditional probability formula is essential because it helps assess scenarios involving dependent events. It allows us to answer questions like "What is the probability of rain given it's cloudy?" and can be extended to more complex situations like Bayesian networks.
Using this formula, one can solve sophisticated problems by breaking them down into parts and understanding how the known affects the unknown. By mastering conditional probability, you're better equipped to handle uncertainty in a structured way.

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Most popular questions from this chapter

An academic department with five faculty membersAnderson, Box, Cox, Cramer, and Fisher-must select two of its members to serve on a personnel review committee. Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting the names on identical pieces of paper and then randomly selecting two. a. What is the probability that both Anderson and Box will be selected? [Hint: List the equally likely outcomes.] b. What is the probability that at least one of the two members whose name begins with \(C\) is selected? c. If the five faculty members have taught for \(3,6,7,10\), and 14 years, respectively, at the university, what is the probability that the two chosen representatives have a total of at least 15 years' teaching experience there?

Fasteners used in aircraft manufacturing are slightly crimped so that they lock enough to avoid loosening during vibration. Suppose that \(95 \%\) of all fasteners pass an initial inspection. Of the \(5 \%\) that fail, \(20 \%\) are so seriously defective that they must be scrapped. The remaining fasteners are sent to a recrimping operation, where \(40 \%\) cannot be salvaged and are discarded. The other \(60 \%\) of these fasteners are corrected by the recrimping process and subsequently pass inspection. a. What is the probability that a randomly selected incoming fastener will pass inspection either initially or after recrimping? b. Given that a fastener passed inspection, what is the probability that it passed the initial inspection and did not need recrimping?

Components of a certain type are shipped to a supplier in batches of ten. Suppose that \(50 \%\) of all such batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. Two components from a batch are randomly selected and tested. What are the probabilities associated with 0,1 , and 2 defective components being in the batch under each of the following conditions? a. Neither tested component is defective. b. One of the two tested components is defective.

At a certain gas station, \(40 \%\) of the customers use regular gas \(\left(A_{1}\right), 35 \%\) use plus gas \(\left(A_{2}\right)\), and \(25 \%\) use premium \(\left(A_{3}\right)\). Of those customers using regular gas, only \(30 \%\) fill their tanks (event \(B\) ). Of those customers using plus, \(60 \%\) fill their tanks, whereas of those using premium, \(50 \%\) fill their tanks. a. What is the probability that the next customer will request plus gas and fill the tank \(\left(A_{2} \cap B\right)\) ? b. What is the probability that the next customer fills the tank? c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?

A family consisting of three persons- \(A, B\), and \(C-\) goes to a medical clinic that always has a doctor at each of stations 1 , 2 , and 3. During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. One outcome is \((1,2,1)\) for \(A\) to station \(1, B\) to station 2 , and \(C\) to station 1 . a. List the 27 outcomes in the sample space. b. List all outcomes in the event that all three members go to the same station. c. List all outcomes in the event that all members go to different stations. d. List all outcomes in the event that no one goes to station 2 .

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