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Chapter 7: Problem 52

Suppose that \(A\) and \(B\) are mutually exclusive events and that \(P(A \cup B) \neq 0\). What is \(P(A \mid A \cup B)\) ?

Short Answer

Expert verified
The probability of event A occurring given the union of events A and B is: \(P(A \mid A \cup B) = \frac{P(A)}{P(A) + P(B)}\).

Step by step solution

01

Definition of Conditional Probability

Recall the definition of the conditional probability: \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) In this case, we want to find \(P(A \mid A \cup B)\), so we will have: \(P(A \mid A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}\)
02

Apply properties of union and intersection to the numerator

Since A is a subset of the union of A and B, the intersection of A and (A 鈭� B) is simply A: \(A \cap (A \cup B) = A\) So, the equation becomes: \(P(A \mid A \cup B) = \frac{P(A)}{P(A \cup B)}\)
03

Use the formula for the probability of the union of two events

The events A and B are mutually exclusive, meaning that their intersection is an empty set (they have no common outcomes). The formula for the probability of the union of two mutually exclusive events is: \(P(A \cup B) = P(A) + P(B)\) Substitute this in the equation from Step 2: \(P(A \mid A \cup B) = \frac{P(A)}{P(A) + P(B)}\)
04

Conclusion

Thus, the probability of event A occurring given the union of events A and B is: \(P(A \mid A \cup B) = \frac{P(A)}{P(A) + P(B)}\)

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Key Concepts

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

Mutually Exclusive Events
Mutually exclusive events are foundational concepts in probability theory. These are events that cannot occur at the same time. In other words, if one event happens, the other cannot. For example, when flipping a fair coin, the event of landing on heads and the event of landing on tails are mutually exclusive鈥攂oth outcomes cannot happen simultaneously on a single coin flip.

The characteristic property of mutually exclusive events is that their intersection (the events occurring together) is an empty set, notated as \( A \cap B = \emptyset \). This property dramatically simplifies calculations in probability, especially when dealing with the probability of unions (the probability that at least one of the events will occur), which we will explore in the next section.
Probability of Union of Events
Understanding the probability of the union of events is pivotal when calculating the chance that at least one of multiple events happens. The probability of the union of two events, \(A\) and \(B\), is denoted as \(P(A \cup B)\).

For mutually exclusive events, such as in the exercise, the probability of their union is the sum of their individual probabilities, giving us \(P(A \cup B) = P(A) + P(B)\). This is because there is no overlap between the events鈥攖here are no outcomes that are part of both \(A\) and \(B\). Consequently, this straightforward addition rule for mutually exclusive events simplifies many probability problems.
Properties of Union and Intersection
The union (\(A \cup B\)) and intersection (\(A \cap B\)) of sets are fundamental operations in set theory and probability. The union of two sets includes all elements that are in \(A\), in \(B\), or in both. In the context of probability, this refers to the likelihood of either event \(A\) occurring, event \(B\) occurring, or any of their shared outcomes occurring, had they not been mutually exclusive.

Intersection of Sets

The intersection refers to elements that are common to both sets. However, for mutually exclusive events, this intersection is null (\(A \cap B = \emptyset\)), meaning there are no common elements between \(A\) and \(B\). Understanding this property is crucial when applying the definition of conditional probability, as it impacts the numerator in the conditional probability formula, simplifying it from \(P(A \cap (A \cup B))\) to \(P(A)\) given that the intersection of \(A\) with the union of \(A\) and \(B\) is merely \(A\) itself.

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Most popular questions from this chapter

A study was conducted among a certain group of union members whose health insurance policies required second opinions prior to surgery. Of those members whose doctors advised them to have surgery, \(20 \%\) were informed by a second doctor that no surgery was needed. Of these, \(70 \%\) took the second doctor's opinion and did not go through with the surgery. Of the members who were advised to have surgery by both doctors, \(95 \%\) went through with the surgery. What is the probability that a union member who had surgery was advised to do so by a second doctor?

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(E\) is an event of an experiment, then \(P(E)+P\left(E^{c}\right)=1\).

In a poll conducted in 2007,2000 adults ages 18 yr and older were asked how frequently they are in touch with their parents by phone. The results of the poll are as follows: $$\begin{array}{lccccc} \hline \text { Answer } & \text { Monthly } & \text { Weekly } & \text { Daily } & \text { Don't know } & \text { Less } \\ \hline \text { Respondents, } \% & 11 & 47 & 32 & 2 & 8 \\ \hline \end{array}$$ If a person who participated in the poll is selected at random, what is the probability that the person said he or she kept in touch with his or her parents a. Once a week? b. At least once a week?

In a survey to determine the opinions of Americans on health insurers, 400 baby boomers and 600 pre-boomers were asked this question: Do you believe that insurers are very responsible for high health costs? Of the baby boomers, 212 answered in the affirmative, whereas 198 of the pre-boomers answered in the affirmative. If a respondent chosen at random from those surveyed answered the question in the affirmative, what is the probability that he or she is a baby boomer? A pre-boomer?

Determine whether the given experiment has a sample space with equally likely outcomes. A ball is selected at random from an urn containing six black balls and six red balls, and the color of the ball is recorded.
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