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Let \(E_{1}, E_{2},\) and \(E_{3}\) be three events from a sample space \(S .\) Find a formula for the probability of \(E_{1} \cup E_{2} \cup E_{3}\) .

Short Answer

Expert verified
Use: \( P(E_{1} \cup E_{2} \cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1} \cap E_{2}) - P(E_{1} \cap E_{3}) - P(E_{2} \cap E_{3}) + P(E_{1} \cap E_{2} \cap E_{3}) \).

Step by step solution

01

Understanding Union of Events

The probability of the union of events, in this case, the events are three: \(E_{1}, E_{2}, \text{ and } E_{3}\). The formula to find the probability of the union of three events is based on the principle of inclusion-exclusion.
02

Applying the Inclusion-Exclusion Principle

The inclusion-exclusion principle states that for any three events \(E_{1}, E_{2}, \text{ and } E_{3}\) from a sample space S, the formula for the probability of their union is: \[ P(E_{1} \cup E_{2} \cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1} \cap E_{2}) - P(E_{1} \cap E_{3}) - P(E_{2} \cap E_{3}) + P(E_{1} \cap E_{2} \cap E_{3}) \].
03

Interpreting the Final Formula

The formula accounts for all possible overlaps between the events: first, we add all individual probabilities, subtract the probabilities of all pairwise intersections (to correct for double-counting), and finally add the probability of the intersection of all three events (since it was subtracted thrice in the pairwise intersections).

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Key Concepts

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

Inclusion-Exclusion Principle
Understanding the Inclusion-Exclusion Principle is essential in Probability Theory. This principle helps us find the probability of the union of several events by considering their intersections. For three events, we need to add their individual probabilities, subtract the probabilities of all pairwise intersections (to adjust for double-counting), and finally, add back the probability of the intersection of all three events. This provides a comprehensive measure of their combined probability. The formula is:
\[ P(E_{1} \cup E_{2} \cup E_{3}) = P(E_{1}) + P(E_{2}) + P(E_{3}) - P(E_{1} \cap E_{2}) - P(E_{1} \cap E_{3}) - P(E_{2} \cap E_{3}) + P(E_{1} \cap E_{2} \cap E_{3}) \].
This ensures we account for all overlaps and avoid counting any event more than once.
Union of Events
The union of events is a core idea in probability. It represents the occurrence of at least one of the events. For events \(E_{1}, E_{2}, \text{ and } E_{3} \) from the same sample space, the union \(E_{1} \cup E_{2} \cup E_{3} \) means that any one of these events happens. The Inclusion-Exclusion Principle aids in calculating this union accurately, especially when the events overlap. To find the probability of the union, we use the formula outlined above, ensuring all intersections are factored in correctly. This method prevents any potential over-counting, providing the precise likelihood of at least one event occurring.
Probability Theory
In Probability Theory, we measure how likely an event is to occur. We express this using a number between 0 and 1, where 0 means the event will not happen, and 1 means it will certainly happen. The principles we use in probability, like the Inclusion-Exclusion Principle, help us deal with complex situations involving multiple events.
When different events can happen together or separately, calculating their combined probability requires careful consideration of their overlaps and intersections. Tools like the Union of Events and the Inclusion-Exclusion Principle are central in these calculations, ensuring our results are accurate and meaningful.
Discrete Mathematics
Discrete Mathematics includes studying structures that are distinct and unconnected, which is crucial in probability problems. It involves combinatorics, set theory, and other topics that help in understanding how to count distinct elements and calculate probabilities properly.
In the context of our problem with events \(E_{1}, E_{2}, \text{ and } E_{3} \), Discrete Mathematics provides the tools to apply the Inclusion-Exclusion Principle effectively.
  • We use set theory to understand unions and intersections of events.
  • Combinatorics helps us count the possible outcomes.
Combining these tools allows us to solve complex probability problems by accurately accounting for all possible overlaps and scenarios.

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Most popular questions from this chapter

Use generating functions to solve the recurrence relation \(a_{k}=4 a_{k-1}-4 a_{k-2}+k^{2}\) with initial conditions \(a_{0}=\) 2 and \(a_{1}=5\) .

If \(G(x)\) is the generating function for the sequence \(\left\\{a_{k}\right\\}\) what is the generating function for each of these sequences? a) \(0,0,0, a_{3}, a_{4}, a_{5}, \ldots\) (assuming that terms follow the pattern of all but the first three terms) b) \(a_{0}, 0, a_{1}, 0, a_{2}, 0, \ldots\) c) \(0,0,0,0, a_{0}, a_{1}, a_{2}, \ldots\) (assuming that terms follow the pattern of all but the first four terms) d) \(a_{0}, 2 a_{1}, 4 a_{2}, 8 a_{3}, 16 a_{4}, \ldots\) e) \(0, a_{0}, a_{1} / 2, a_{2} / 3, a_{3} / 4, \ldots[\text { Hint: Calculus required }\) here. \(]\) f) \(a_{0}, a_{0}+a_{1}, a_{0}+a_{1}+a_{2}, a_{0}+a_{1}+a_{2}+a_{3}, \ldots\)

Generating functions are useful in studying the number of different types of partitions of an integer \(n .\) A partition of a positive integer is a way to write this integer as the sum of positive integers where repetition is allowed and the or- der of the integers in the sum does not matter. For exam- ple, the partitions of 5 (with no restrictions) are \(1+1+1+1+\) \(1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,\) and \(5 .\) Exercises \(53-58\) illustrate some of these uses. Show that the coefficient \(p_{d}(n)\) of \(x^{n}\) in the formal power series expansion of \((1+x)\left(1+x^{2}\right)\left(1+x^{3}\right) \cdots\) equals the number of partitions of \(n\) into distinct parts, that is, the number of ways to write \(n\) as the sum of positive inte- gers, where the order does not matter but no repetitions are allowed.

Generating functions are useful in studying the number of different types of partitions of an integer \(n .\) A partition of a positive integer is a way to write this integer as the sum of positive integers where repetition is allowed and the or- der of the integers in the sum does not matter. For exam- ple, the partitions of 5 (with no restrictions) are \(1+1+1+1+\) \(1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,\) and \(5 .\) Exercises \(53-58\) illustrate some of these uses. Find \(p_{o}(n),\) the number of partitions of \(n\) into odd parts with repetitions allowed, and \(p_{d}(n),\) the number of par- titions of \(n\) into distinct parts, for \(1 \leq n \leq 8,\) by writing each partition of each type for each integer.

In the Tower of Hanoi puzzle, suppose our goal is to transfer all \(n\) disks from peg 1 to peg \(3,\) but we cannot move a disk directly between pegs 1 and \(3 .\) Each move of a disk must be a move involving peg \(2 .\) As usual, we cannot place a disk on top of a smaller disk. a) Find a recurrence relation for the number of moves required to solve the puzzle for \(n\) disks with this added restriction. b) Solve this recurrence relation to find a formula for the number of moves required to solve the puzzle for \(n\) disks. c) How many different arrangements are there of the \(n\) disks on three pegs so that no disk is on top of a smaller disk? d) Show that every allowable arrangement of the \(n\) disks occurs in the solution of this variation of the puzzle.

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