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Chapter 2: Problem 104

If \(A\) and \(B\) are two events, prove that \(P(A \cap B) \geq 1-P(\bar{A})-P(\bar{B})\). \([\) Note: This is a simplified version of the Bonferroni inequality.]

Short Answer

Expert verified
Hence, we have proved \(P(A \cap B) \geq 1-P(\bar{A})-P(\bar{B})\).

Step by step solution

01

Express the Complement of Events

To solve the inequality, let's first express the complement of events. The complement of event \(A\) is given by \(\bar{A}\), and similarly, the complement of event \(B\) is \(\bar{B}\).
02

Apply Probability Addition Rule

Recall the probability addition rule for two events: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). This can be used to express \(P(A \cap B)\) in terms of \(P(A \cup B)\).
03

Relate Complements to Union

Since \(\bar{A}\cap\bar{B}\) is the complement of \(A\cup B\), we have \(P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B)\).
04

Use Probability of Complements

By the property of complements for individual events, \(P(\bar{A} \cup \bar{B}) = P(\bar{A}) + P(\bar{B}) - P(\bar{A} \cap \bar{B})\).
05

Simplify the Relation

Now, using steps 3 and 4, we have the equation \(P(\bar{A}) + P(\bar{B}) - P(\bar{A} \cap \bar{B}) = P(\bar{A}) + P(\bar{B}) - (1-P(A \cup B))\). Rearrange it to find \(P(A \cup B) \geq 1 - P(\bar{A}) - P(\bar{B})\).
06

Apply Inequality to Intersection

Rewriting the equation in terms of \(P(A \cap B)\), we know: \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\). Since \(P(A \cup B) \geq 1 - P(\bar{A}) - P(\bar{B})\), we rearrange to show \(P(A \cap B) \geq 1 - P(\bar{A}) - P(\bar{B})\).

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics dealing with the likelihood of events occurring. It is fundamental for understanding how and why certain outcomes happen. In probability theory, we talk about events, their complements, unions, and intersections.
  • **Events** are outcomes or occurrences that can happen in an experiment.
  • **Probabilities** are numbers between 0 and 1 that represent the likelihood of these events.
  • A probability of 0 means the event will not happen, whereas a probability of 1 means it is certain to occur.
Probability theory uses principles and formulas to calculate and predict the chances of different outcomes.
When analyzing situations, probabilities can help us understand complex phenomena and make informed decisions. These principles are essential when dealing with conjunctions, disjunctions, and complementary occurrences of events, providing the foundation for more advanced concepts like the Bonferroni inequality.
Complement of an Event
In probability, the complement of an event refers to the outcomes in the sample space that are not part of the original event. For any event \(A\), its complement is denoted by \(\bar{A}\) or sometimes \(A^c\).
The complement rule states that the probability of an event plus the probability of its complement equals 1:
  • \(P(A) + P(\bar{A}) = 1\)
  • Therefore, \(P(\bar{A}) = 1 - P(A)\).
Understanding complements is crucial when applying rules like the Bonferroni inequality, as often restating a problem in terms of complements can simplify it significantly.
In practice, if you know the probability of something not happening, you automatically know the probability of it happening. This can be particularly useful in calculating probabilities involving unions and intersections of events.
Intersection of Events
The intersection of events occurs when two or more events happen simultaneously. If events \(A\) and \(B\) intersect, it means both event \(A\) and event \(B\) have occurred. The intersection is denoted as \(A \cap B\).
The probability of the intersection is calculated using the formula:
  • \(P(A \cap B) = P(A) + P(B) - P(A \cup B)\)
This demonstrates the probability that both events \(A\) and \(B\) will occur. Understanding the concept of intersection helps in determining how jointly probable two events are, which is vital when applying more complex equations like those related to the Bonferroni inequality.
Furthermore, analyzing intersections can assist in risk assessments and decision-making processes, as many real-world situations require knowing how multiple factors work together.

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Most popular questions from this chapter

An advertising agency notices that approximately 1 in 50 potential buyers of a product sees a given magazine ad, and 1 in 5 sees a corresponding ad on television. One in 100 sees both. One in 3 actually purchases the product after seeing the ad, 1 in 10 without seeing it. What is the probability that a randomly selected potential customer will purchase the product?

Two methods, \(A\) and \(B\), are available for teaching a certain industrial skill. The failure rate is \(20 \%\) for \(A\) and \(10 \%\) for \(B\). However, \(B\) is more expensive and hence is used only \(30 \%\) of the time. \((A\) is used the other \(70 \% .\) ) A worker was taught the skill by one of the methods but failed to learn it correctly. What is the probability that she was taught by method \(A\) ?

Suppose that two defective refrigerators have been included in a shipment of six refrigerators. The buyer begins to test the six refrigerators one at a time. a. What is the probability that the last defective refrigerator is found on the fourth test? b. What is the probability that no more than four refrigerators need to be tested to locate both of the defective refrigerators? c. When given that exactly one of the two defective refrigerators has been located in the first two tests, what is the probability that the remaining defective refrigerator is found in the third or fourth test?

Five cards are dealt from a standard 52 -card deck. What is the probability that we draw a. 1 ace. 1 two, 1 three, 1 four, and 1 five (this is one way to get a "straight")? b. any straight?

A sample space consists of five simple events, \(E_{1}, E_{2}, E_{3}, E_{4},\) and \(E_{5}\) a. If \(P\left(E_{1}\right)=P\left(E_{2}\right)=0.15, P\left(E_{3}\right)=0.4,\) and \(P\left(E_{4}\right)=2 P\left(E_{5}\right),\) find the probabilities of \(E_{4}\) and \(E_{5}\) b. If \(P\left(E_{1}\right)=3 P\left(E_{2}\right)=0.3,\) find the probabilities of the remaining simple events if you know that the remaining simple events are equally probable.
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