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If \(A\) and \(B\) are mutually exclusive events and \(P(B)>0\), show that $$P(A | A \cup B)=\frac{P(A)}{P(A)+P(B)}$$.

Short Answer

Expert verified
The conditional probability is \( P(A | A \cup B) = \frac{P(A)}{P(A)+P(B)} \).

Step by step solution

01

Understand the Problem

In this problem, we are given two mutually exclusive events, \(A\) and \(B\). This means \(P(A \cap B) = 0\), as they cannot occur simultaneously. We need to find the conditional probability \(P(A | A \cup B)\).
02

Apply the Conditional Probability Formula

To find \(P(A | A \cup B)\), we use the conditional probability formula \[ P(A | A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}. \] Note that the event \(A \cap (A \cup B)\) is simply \(A\), since \(A\) is part of the union \(A \cup B\). Thus, \(P(A \cap (A \cup B)) = P(A)\).
03

Find the Probability of the Union

The formula for the probability of the union of two events is given by \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Since \(A\) and \(B\) are mutually exclusive, \(P(A \cap B) = 0\). Thus, \(P(A \cup B) = P(A) + P(B)\).
04

Substitute and Simplify

Substitute the results from Steps 2 and 3 into the conditional probability formula: \[ P(A | A \cup B) = \frac{P(A)}{P(A) + P(B)}. \] This is the required expression for the conditional probability.

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Key Concepts

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

Mutually Exclusive Events
In probability theory, two events are called mutually exclusive if they cannot happen at the same time. This is an important concept when calculating probabilities, as it can simplify the process significantly. For instance, in the exercise provided, events \(A\) and \(B\) are mutually exclusive. This means that the occurrence of event \(A\) rules out the occurrence of event \(B\) and vice versa.

When two events \(A\) and \(B\) are mutually exclusive, their intersection is empty. Thus, the probability of both events occurring simultaneously is zero: \(P(A \cap B) = 0\).
  • Two events with no common outcomes cannot happen at the same time.
  • The concept is crucial in simplifying how we find the probability of combinations of these events.
  • It's important to recognize mutually exclusive events in questions, as it impacts calculations like the probability of their union.
In practical examples, think about flipping a coin: the outcomes "heads" and "tails" are mutually exclusive because you cannot get both at the same time. Understanding this helps make sense of their collective probabilities.
Probability Union
Calculating the probability of either event \(A\) or event \(B\) happening, known as the probability of their union, involves considering whether events \(A\) and \(B\) are mutually exclusive. This is because their exclusivity informs the formula to use.

For any events \(A\) and \(B\), the probability union is found with the formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). The term \(P(A \cap B)\) adjusts for any overlap between the two events, since both events occurring together should not be counted twice.
  • If \(A\) and \(B\) are mutually exclusive, then \(P(A \cap B) = 0\), significantly simplifying the calculation to \(P(A \cup B) = P(A) + P(B)\).
  • This formula highlights the integral role that mutual exclusivity plays in probability.
  • The union probability shows the likelihood of at least one of the events occuring.
In our initial exercise, because \(A\) and \(B\) are mutually exclusive, this simplification made finding \(P(A \cup B)\) easier, leading us to a clear path for determining conditional probabilities.
Conditional Probability Formula
The conditional probability formula allows us to find the likelihood of event \(A\) occurring given that some related event (like \(A \cup B\)) already occurs. It provides a focused perspective by narrowing down the sample space to situations where the given condition happens.

The formula for conditional probability, \(P(A | B)\), is expressed as \(P(A | B) = \frac{P(A \cap B)}{P(B)}\). This reads as "the probability of \(A\) given \(B\) is equal to the probability of both \(A\) and \(B\) happening, divided by the probability of \(B\)."
  • It highlights the chance of an event given a specific condition is met.
  • In the exercise, \(P(A | A \cup B)\) simplifies using the properties of mutually exclusive events.
  • This makes the concepts easier to manage, as certain probabilities zero out.
Understanding how to correctly apply the conditional probability formula is vital when solving problems involving linked events, especially when interested in the likelihood relative to another known occurrence.

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Most popular questions from this chapter

If \(A_{1}, A_{2},\) and \(A_{3}\) are three events and \(P\left(A_{1} \cap A_{2}\right)=P\left(A_{1} \cap A_{3}\right) \neq 0\) but \(P\left(A_{2} \cap A_{3}\right)=0,\) show that \(P\left(\text { at least one } A_{i}\right)=P\left(A_{1}\right)+P\left(A_{2}\right)+P\left(A_{3}\right)-2 P\left(A_{1} \cap A_{2}\right)\)

A personnel director has two lists of applicants for jobs. List 1 contains the names of five women and two men, whereas list 2 contains the names of two women and six men. A name is randomly selected from list 1 and added to list \(2 .\) A name is then randomly selected from the augmented list 2\. Given that the name selected is that of a man, what is the probability that a woman's name was originally selected from list \(1 ?\)

An experimenter wishes to investigate the effect of three variables \(-\) pressure, temperature, and the type of catalyst-on the yield in a refining process. If the experimenter intends to use three settings each for temperature and pressure and two types of catalysts, how many experimental runs will have to be conducted if he wishes to run all possible combinations of pressure, temperature, and types of catalysts?

Five firms, \(F_{1}, F_{2}, \ldots, F_{5},\) each offer bids on three separate contracts, \(C_{1}, C_{2},\) and \(C_{3} .\) Any one firm will be awarded at most one contract. The contracts are quite different, so an assignment of \(C_{1}\) to \(F_{1},\) say, is to be distinguished from an assignment of \(C_{2}\) to \(F_{1}\). a. How many sample points are there altogether in this experiment involving assignment of contracts to the firms? (No need to list them all.) b. Under the assumption of equally likely sample points, find the probability that \(F_{3}\) is awarded a contract.

A group of five applicants for a pair of identical jobs consists of three men and two women. The employer is to select two of the five applicants for the jobs. Let \(S\) denote the set of all possible outcomes for the employer's selection. Let \(A\) denote the subset of outcomes corresponding to the selection of two men and \(B\) the subset corresponding to the selection of at least one woman. List the outcomes in \(A, \bar{B}, A \cup B, A \cap B,\) and \(A \cap \bar{B}\). (Denote the different men and women by \(M_{1}, M_{2}, M_{3}\) and \(W_{1}, W_{2},\) respectively.)

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