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Let \(0

Short Answer

Expert verified
The equation is confirmed using conditional probabilities and the law of total probability.

Step by step solution

01

Understanding Conditional Probability

Conditional probability, denoted as \( P(A \mid B) \), is the probability of event \( A \) occurring given that \( B \) has occurred. The formula for \( P(A \mid B) \) is: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \] Similarly, \( P(A \mid B^c) \) is \[ P(A \mid B^c) = \frac{P(A \cap B^c)}{P(B^c)} \].
02

Expressing P(A) in terms of intersections

The probability of event \( A \) can be expressed in terms of the mutually exclusive events \( A \cap B \) and \( A \cap B^c \). Since these two intersections cover all possibilities for \( A \) (whether \( B \) occurs or not), we have: \[ P(A) = P(A \cap B) + P(A \cap B^c) \].
03

Using Conditional Probability Definitions

Using the definitions of conditional probability, we can express \( P(A \cap B) \) as \( P(A \mid B) P(B) \) and \( P(A \cap B^c) \) as \( P(A \mid B^c) P(B^c) \). Substitute these into the equation from Step 2: \[ P(A) = P(A \mid B) P(B) + P(A \mid B^c) P(B^c) \].
04

Conclusion

We have shown that \( P(A) \) can be decomposed into its conditional probabilities and the probabilities of the conditions themselves, confirming the given equation.

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Key Concepts

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

Conditional Probability
Conditional probability, often denoted as \( P(A \mid B) \), is a foundational concept in probability theory. It allows you to calculate the probability of an event \( A \) happening, given that another event \( B \) has already occurred. Essentially, it's a measure of how likely one event is, after taking into account the occurrence of a related event.
For instance, the equation \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \) illustrates how conditional probability is derived. You look at the intersection of \( A \) and \( B \) (both events happening together), divided by the probability of \( B \) occurring.
This method is crucial when dealing with sequences of events where conditions affect outcomes. It simplifies cases where assuming independence would be unrealistic or impossible.
Probability Theory
Probability theory is the mathematical framework used to study the likelihood of events. It provides tools and concepts to quantify uncertain outcomes. With probability, we can make predictions and assess risks in various domains such as finance, science, and everyday life.
  • **Random Variables:** Functions that assign a numerical outcome to an event. Examples include the roll of a die or the result of a coin flip.
  • **Probability Space:** A model consisting of \( S \), the sample space of all possible outcomes, \( \mathcal{F} \), the set of all possible events, and \( P \), the probability measure that assigns a probability to each event.

Fundamental to probability theory is ensuring that probabilities are within the bounds of 0 and 1, where 0 means impossibility and 1 means certainty. Additionally, the sum of probabilities for all possible outcomes must equal 1, reflecting the certainty that something will occur.
Events and Outcomes
In probability, an event is any outcome or set of outcomes of a random experiment. Understanding the distinction between events and outcomes is critical to solving probabilistic problems.
  • **Outcome:** A single possible result of an experiment, such as rolling a specific number on a die.
  • **Event:** A specific set of outcomes we're interested in. For instance, rolling an even number is an event consisting of the outcomes 2, 4, and 6.

Events can be simple or compound, where compound events combine two or more simple events. By understanding these basic elements—events and outcomes—you can analyze and calculate probabilities more effectively.

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Most popular questions from this chapter

Suppose \(A, B, C\) are independent events and \(P(A \cap B) \neq 0\). Show \(P(C \mid A \cap B)=P(C)\).

An insurance company insures an equal number of male and female drivers. In any given year the probability that a male driver has an accident involving a claim is \(\alpha\), independently of other years. The analogous probability for females is \(\beta\). Assume the insurance company selects a driver at random. a) What is the probability the selected driver will make a claim this year? \(\left[\right.\) Ans:: \(\left.\frac{\alpha+\beta}{2}\right]\) b) What is the probability the selected driver makes a claim in two consecutive years? \(\left[\right.\) Ans.. \(\left.\frac{\alpha^{2}+\beta^{2}}{2}\right]\)

In all exercises the probability space is fixed, and \(A . B, A_{n}\), etc... are events. Show that if \(A \cap B=\emptyset\), then \(A\) and \(B\) cannot be independent unless \(P(A)=0\) or \(P(B)=0\).

Suppose \(P(C)>0\) and \(A_{1} \ldots . A_{n}\) are all pairwise disjoint. Show that $$ P\left(\cup_{i=1}^{n} A_{i} \mid C\right)=\sum_{i=1}^{n} P\left(A_{i} \mid C\right) . $$

A box has \(r\) red and \(b\) black balls. A ball is chosen at random from the box (so that each ball is equally likely to be chosen), and then a second ball is drawn at random from the remaining balls in the box. Find the probabilities that a) Both balls are red \(\left[\right.\) Ans: \(\left.\frac{r(r-1)}{(r+b)(r+b-1)}\right]\) b) The first ball is red and the second is black [Ans.: \(\left.\frac{r b}{(r+b)(r+b-1)}\right]\)

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