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91Ó°ÊÓ

Answer questions true or false and give a brief explanation for each answer. Hint: Review the summary of basic probability rules.\(P(A \mid B)+P\left(A^{c} \mid B\right)=1\)

Short Answer

Expert verified
True. The expression uses the rule of complementary probabilities under condition B.

Step by step solution

01

Understanding Conditional Probability

Conditional probability represents the probability of an event occurring given that another event has already occurred. In this case, we are looking at events occurring under the condition event B has occurred.
02

Basics of Complementary Events

Complementary events are two outcomes of an event that sum up to 1. Essentially, if event A occurs, its complement, denoted as \(A^c\), does not occur, and vice versa.
03

Applying Complementary Rule to Conditional Probability

Given the expression \(P(A \mid B) + P(A^{c} \mid B) = 1\), it applies the complementary probability rule. For any event A and its complement \(A^c\), the sum of their probabilities should equal 1, even under the condition of event B.
04

Evaluating the Given Expression

The expression states that the probability of A occurring given B, plus the probability of A not occurring given B, equals 1. This matches the rule of complementary events under the condition B, validating the expression.

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Key Concepts

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

Probability Rules
Probability rules are fundamental principles used to calculate the likelihood of events. Understanding these rules helps in solving complex probability problems, including those involving conditional probabilities. One key rule in probability is the addition rule for complementary events. When considering any event A, the probability of A per se, denoted as \( P(A) \), plus the probability of its complement \( A^c \), i.e., \( P(A^c) \), must equal 1 because an event either happens or it doesn't.
Another important probability rule is the multiplication rule used when there's a need to find joint probabilities. This often applies when determining the probability of two independent events, A and B, happening together, denoted as \( P(A \cap B) = P(A) \times P(B) \) when they are independent. Understanding these rules helps you dissect problems to find solutions effectively.
  • The addition rule: applies to mutually exclusive events, ensuring they sum up to a total probability of 1.
  • The multiplication rule: used for finding joint probabilities of independent events.
Combining these rules with an understanding of complementary events helps solve conditional probability problems accurately.
Complementary Events
Complementary events refer to a pair of mutually exclusive outcomes where the occurrence of one implies that the other does not occur. The probability of an event A and its complementary event \( A^c \) sums to 1. This principle is especially useful when the probability of an event is easier to calculate through its complement.
Consider a situation where you are determining the probability of not rolling a 6 on a six-sided die. The event of rolling a 6 is denoted as A, and its complement, not rolling a 6, is \( A^c \). Since there is only one side with a 6, the probability \( P(A) = \frac{1}{6} \). Thus, \( P(A^c) = 1 - \frac{1}{6} = \frac{5}{6} \).
  • Complementary events are a useful concept because if you know \( P(A) \), you can easily find \( P(A^c) = 1 - P(A) \).
  • Understanding and identifying complementary events can simplify calculation and provide clarity in probability theory.
It's essential in probability theory, especially with conditional probabilities, to apply this principle and ensure total probabilities align correctly.
Probability Theory
Probability theory is the branch of mathematics that deals with calculating the likelihood of different outcomes. It lays the groundwork for statistics and helps in predicting outcomes in uncertain situations. Conditional probability is an essential part of this theory, representing the probability of an event occurring given that another event has already occurred.
In the context of conditional probability, an important formula involves complementary events: \( P(A \mid B) + P(A^c \mid B) = 1 \). This is a reflection of the complementary principle within the framework of conditional probability. It states that even when you're considering events based on a condition, their probabilities should conform to established probability laws.
  • Probability theory combines elements like probability distributions, random variables, and expected values to explain outcomes.
  • Conditional probability focuses on the impact of one event on another, a key aspect of the theory requiring a solid understanding of dependency and independence between events.
By mastering these probability concepts, you're well-equipped to tackle questions related to risk, decision-making, and predictive analyses.

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Most popular questions from this chapter

Given \(P(A)=0.3\) and \(P(B)=0.4\) : (a) If \(A\) and \(B\) are mutually exclusive events, compute \(P(A\) or \(B)\). (b) If \(P(A\) and \(B)=0.1\), compute \(P(A\) or \(B)\).

Given \(P(A)=0.2 ; P(B)=0.5\), \(P(A \mid B)=0.3:\) (a) Compute \(P(A\) and \(B)\). (b) Compute \(P(A\) or \(B)\).

Alcobol Recovery The Eastmore Program is a special program to help alcoholics. In the Eastmore Program, an alcoholic lives at home but undergoes a two-phase treatment plan. Phase I is an intensive group-therapy program lasting 10 weeks. Phase II is a long-term counseling program lasting 1 year. Eastmore Programs are located in most major cities, and past data gave the following information based on percentages of success and failure collected over a long period of time: The probability that a client will have a relapse in phase I is \(0.27\); the probability that a client will have a relapse in phase II is \(0.23\). However, if a client did not have a relapse in phase I, then the probability that this client will not have a relapse in phase II is \(0.95\). If a client did have a relapse in phase I, then the probability that this client will have a relapse in phase II is \(0.70\). Let \(A\) be the event that a client has a relapse in phase I and \(B\) be the event that a client has a relapse in phase II. Let \(C\) be the event that a client has no relapse in phase \(\mathrm{I}\) and \(D\) be the event that a client has no relapse in phase II. Compute the following: (a) \(P(A), P(B), P(C)\), and \(P(D)\) (b) \(P(B \mid A)\) and \(P(D \mid C)\) (c) \(P(A\) and \(B)\) and \(P(C\) and \(D)\) (d) \(P(A\) or \(B)\) (e) What is the probability that a client will go through both phase I and phase II without a relapse? (f) What is the probability that a client will have a relapse in both phase I and phase II? (g) What is the probability that a client will have a relapse in either phase I or phase II?

If two events are mutually exclusive, can they occur concurrently? Explain.

Answer questions true or false and give a brief explanation for each answer. Hint: Review the summary of basic probability rules.$$ P\left(A^{c} \text { and } B^{c}\right) \leq 1-P(A) $$

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