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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 identity is proven using the additivity of probabilities and the definition of conditional probability.

Step by step solution

01

Understand Conditional Probability

The formula for conditional probability is given by \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \). Similarly, \( P(A' \mid B) = \frac{P(A' \cap B)}{P(B)} \), where \( A' \) is the complement of \( A \).
02

Intersection of Complements

The complement of an event \( A \), denoted \( A' \), includes all outcomes not in \( A \). Thus, \( A \cap B \) and \( A' \cap B \) are mutually exclusive and exhaustive for the event \( B \). This implies: \( P(A \cap B) + P(A' \cap B) = P(B) \).
03

Substitute into Conditional Probability

Using the result from the previous step, substitute into the conditional probabilities: \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \) and \( P(A' \mid B) = \frac{P(A' \cap B)}{P(B)} \).
04

Add Conditional Probabilities Together

Add the conditional probabilities: \[ P(A \mid B) + P(A' \mid B) = \frac{P(A \cap B)}{P(B)} + \frac{P(A' \cap B)}{P(B)} \].
05

Simplify the Expression

Combine the fractions: \[ \frac{P(A \cap B) + P(A' \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1 \]. This shows that \( P(A \mid B) + P(A' \mid B) = 1 \).

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Key Concepts

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

Complementary Events
In the world of probability, complementary events play a crucial role. Imagine an event, say event A happening when you toss a coin and it shows heads. The complement of this event, noted as \( A' \), would then be the occurrence where the coin shows tails. Simply put, complementary events \( A \) and \( A' \) entirely cover all possible outcomes of the sample space.

This notion is essential, as it sets up scenarios where the probability of \( A \) or \( A' \) occurring sums to 1. Mathematically, we express this as:
  • \( P(A) + P(A') = 1 \)
When you look at conditional probability, complementary events look a bit different. Given events \( A \) and \( B \), where \( B \) is in some context (like only considering rainy days), the probability \( P(A \mid B) \) and \( P(A' \mid B) \), still sum to 1. This tells us the likelihood of event A or its complement happening, given B has already occurred.

This understanding of complementary events is fundamental in assessing situations fully, knowing that together, an event and its complement account for every possible outcome within the confined space of event \( B \).
Mutually Exclusive Events
Mutually exclusive events are fascinating in probability because they describe a scenario where two events, say \( A \) and \( B \), cannot happen at the same time. If you rolled a die, getting a 1 and a 2 simultaneously is impossible, thus, these events are mutually exclusive.

In terms of probability, for mutually exclusive events \( A \) and \( B \), the probability of both events occurring at the same time, noted \( P(A \cap B) \), is zero. This property is important when calculating overall probabilities:
  • \( P(A \cup B) = P(A) + P(B) \)
However, in the context of conditional probability, while direct mutual exclusivity between \( A \) and \( B \) might not always apply, for \( A \cap B \) and \( A' \cap B \), these events are seen as mutually exclusive. This means within the conditions set by \( B \), \( A \) and its complement \( A' \) define non-overlapping possibilities.

Understanding this helps analyze complex situations by noting when combined events are mutually exclusive or not, and how their interplay affects probability distributions.
Probability Axioms
Probability axioms provide the foundational rules upon which all probability theory is built. These axioms are like the grammar rules that keep the chaos of probability in check.

There are three main axioms:
  • **Non-negativity**: For any event \( A \), the probability of \( A \), \( P(A) \), must be greater than or equal to zero.
  • **Normalization**: The probability of the whole sample space is 1. If you're certain that some event will happen, that event has a probability of 1.
  • **Additivity**: For any two mutually exclusive events \( A \) and \( B \), the probability of either \( A \) or \( B \) occurring is the sum of their individual probabilities.
In the case of conditional probability, these axioms still hold true, modified to reflect conditions applied to certain subsets. Understanding these axioms lays the groundwork for grasping more specific probability scenarios, including where conditional relations and complements play a key role.

These axioms guarantee that any calculation of probabilities stays consistent and reliable, making them essential for accurately determining the likelihood of events occurring, whether you're looking at single events or analyzing conditional probabilities.

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Most popular questions from this chapter

1,30 \%\( of the time on airline \)\\#… # A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; \(50 \%\) of the time she travels on airline \(\\# 1,30 \%\) of the time on airline \(\\# 2\), and the remaining \(20 \%\) of the time on airline #3. For airline #1, flights are late into D.C. \(30 \%\) of the time and late into L.A. \(10 \%\) of the time. For airline \(\\# 2\), these percentages are \(25 \%\) and \(20 \%\), whereas for airline \(\\# 3\) the percentages are \(40 \%\) and \(25 \%\). If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines \(\\# 1\), #2, and #3? Assume that the chance of a late arrival in L.A. is unaffected by what happens on the flight to D.C.

A box in a supply room contains 15 compact fluorescent lightbulbs, of which 5 are rated 13-watt, 6 are rated 18-watt, and 4 are rated 23-watt. Suppose that three of these bulbs are randomly selected. a. What is the probability that exactly two of the selected bulbs are rated 23-watt? b. What is the probability that all three of the bulbs have the same rating? c. What is the probability that one bulb of each type is selected? d. If bulbs are selected one by one until a 23-watt bulb is obtained, what is the probability that it is necessary to examine at least 6 bulbs?

Show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B)\), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=\) \(A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B), P(A)\) and \(P(A \cup B)\) ?

Consider the following information about travelers on vacation (based partly on a recent Travelocity poll): \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them, \(23 \%\) both check work email and use a cell phone to stay connected, and \(51 \%\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition, 88 out of every 100 who bring a laptop also check work email, and 70 out of every 100 who use a cell phone to stay connected also bring a laptop. a. What is the probability that a randomly sclccted traveler who checks work email also uses a cell phone to stay connected? b. What is the probability that someone who brings a laptop on vacation also uses a cell phone to stay connected? c. If the randomly selected traveler checked work email and brought a laptop, what is the probability that he/ she uses a cell phone to stay connected?

Suppose that \(55 \%\) of all adults regularly consume coffee, \(45 \%\) regularly consume carbonated soda, and \(70 \%\) regularly consume at least one of these two products. a. What is the probability that a randomly selected adult regularly consumes both coffee and soda? b. What is the probability that a randomly selected adult doesn't regularly consume at least one of these two products?

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