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If \(A, B,\) and \(C\) are any three events, show that $$ \begin{aligned} P(A \cup B \cup C)=& P(A)+P(B)+P(C) \\ &-P(A \cap B)-P(B \cap C)-P(C \cap A) \\ &+P(A \cap B \cap C) \end{aligned} $$

Short Answer

Expert verified
The probability of the union of three events is given by the Inclusion-Exclusion Principle.

Step by step solution

01

Understand Union Probability

We want to find the probability that at least one of the events \(A\), \(B\), or \(C\) occurs. This is given by \(P(A \cup B \cup C)\). To calculate this, we need to use the Inclusion-Exclusion Principle, which helps us handle overlaps between these events.
02

Apply Probability Addition Rule

Start by adding the probabilities of each event separately: \(P(A) + P(B) + P(C)\). This sum accounts for all three events individually but counts each overlap twice.
03

Subtract Pairwise Intersections

To correct for counting overlaps twice, subtract the probabilities of all pairwise intersections: \(P(A \cap B)\), \(P(B \cap C)\), and \(P(C \cap A)\). Each intersection represents a double-counted area that we need to remove once.
04

Add Back Triple Intersection

The intersection of all three events, \(P(A \cap B \cap C)\), has been subtracted three times when eliminating pairwise overlaps, so it must be added back once to account for this triple intersection presence correctly.
05

Compile the Formula

By putting together all these adjusted parts, the probability of the union of the three events is given by the formula: \[P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C).\]

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Key Concepts

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

Union Probability
In probability theory, the term \(P(A \cup B \cup C)\) refers to the probability that at least one of the events \(A\), \(B\), or \(C\) occurs. This is called the union probability of the events. When calculating this probability, one needs to account for the possibility of overlap between the events.
To visualize this, imagine each event as a circle in a Venn diagram. The probability of the union covers all areas within these circles. Using the Inclusion-Exclusion Principle is crucial here to ensure correct calculation, especially because events may overlap or occur simultaneously. Identifying intersections will be key to determining the union probability accurately.
Each step in this calculation ensures that we appropriately include all events and correct for overlaps. This aim is to determine how likely it is for at least one of these events to happen.
Probability Addition Rule
The Probability Addition Rule helps us start our calculation process for union probability by considering each event's individual probabilities. Initially, we add these probabilities together: \(P(A) + P(B) + P(C)\). This first sum implies that each event is independent of overlaps, hence it sums up each circle's area in the Venn diagram.
This basic step assumes no interaction between events, meaning that each event is calculated as if it's unrelated to the others. At this point, event overlaps are not yet considered; thus, they are potentially counted multiple times if they exist.
Understanding this addition is our starting point, but it doesn't give an entirely accurate result until we handle intersections in the subsequent steps.
Pairwise Intersections
To correct the over-counting from the Probability Addition Rule, we must subtract the probability of pairwise intersections: \(P(A \cap B)\), \(P(B \cap C)\), and \(P(C \cap A)\). Each of these terms corresponds to where two events intersect, meaning they happen at the same time. In a Venn diagram, these are the areas where two circles overlap.
The subtraction acknowledges that the overlap regions have been counted twice so far—once with each event's probability. By removing these double-counted portions, we refine our union probability calculation.
For instance, \(P(A \cap B)\) excludes the count of both \(A\) and \(B\) when they simultaneously occur. It reflects the reality of events intersecting in practical scenarios. This precise adjustment ensures overlaps aren't wrongly inflated in our probability estimate.
Triple Intersection
Finally, we must address the situation where all three events, \(A\), \(B\), and \(C\), occur simultaneously, represented by \(P(A \cap B \cap C)\). When subtracting pairwise intersections previously, we inadvertently subtracted the triple intersection more than once.
Adding back \(P(A \cap B \cap C)\) corrects this oversight. It ensures that the area where all three circles overlap on a Venn diagram is counted just once in the final probability.
This step reflects real-world instances where all events occur together. It verifies that this unique scenario of full overlap is accounted for appropriately. Thus, the triple intersection ensures our calculations precisely depict the actual likelihood across all considered events.

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Most popular questions from this chapter

(a) For events \(A_{1}, \ldots, A_{n},\) prove that $$ P\left(A_{1} \cup \cdots \cup A_{n}\right) \leq P\left(A_{1}\right)+\cdots+P\left(A_{n}\right) $$ (b) For events \(A\) and \(B\), prove that $$ P(A \cap B) \geq P(A)+P(B)-1 $$

Let \(X\) be a random variable with distribution function \(m_{X}(x)\) defined by $$ m_{X}(-1)=1 / 5, \quad m_{X}(0)=1 / 5, \quad m_{X}(1)=2 / 5, \quad m_{X}(2)=1 / 5 $$ (a) Let \(Y\) be the random variable defined by the equation \(Y=X+3\). Find the distribution function \(m_{Y}(y)\) of \(Y\). (b) Let \(Z\) be the random variable defined by the equation \(Z=X^{2}\). Find the distribution function \(m_{Z}(z)\) of \(Z\).

Consider the bet that all three dice will turn up sixes at least once in \(n\) rolls of three dice. Calculate \(f(n)\), the probability of at least one triple- six when three dice are rolled \(n\) times. Determine the smallest value of \(n\) necessary for a favorable bet that a triple-six will occur when three dice are rolled \(n\) times. (DeMoivre would say it should be about \(216 \log 2=149.7\) and so would answer 150 - see Exercise 1.2.17. Do you agree with him?)

Here is an attempt to get around the fact that we cannot choose a "random integer." (a) What, intuitively, is the probability that a "randomly chosen" positive integer is a multiple of \(3 ?\) (b) Let \(P_{3}(N)\) be the probability that an integer, chosen at random between 1 and \(N,\) is a multiple of 3 (since the sample space is finite, this is a legitimate probability). Show that the limit $$ P_{3}=\lim _{N \rightarrow \infty} P_{3}(N) $$ exists and equals \(1 / 3\). This formalizes the intuition in (a), and gives us a way to assign "probabilities" to certain events that are infinite subsets of the positive integers. (c) If \(A\) is any set of positive integers, let \(A(N)\) mean the number of elements of \(A\) which are less than or equal to \(N\). Then define the "probability" of \(A\) as $$ P(A)=\lim _{N \rightarrow \infty} A(N) / N $$ provided this limit exists. Show that this definition would assign probability 0 to any finite set and probability 1 to the set of all positive integers. Thus, the probability of the set of all integers is not the sum of the probabilities of the individual integers in this set. This means that the definition of probability given here is not a completely satisfactory definition. (d) Let \(A\) be the set of all positive integers with an odd number of digits. Show that \(P(A)\) does not exist. This shows that under the above definition of probability, not all sets have probabilities.

Tversky and Kahneman \(^{23}\) asked a group of subjects to carry out the following task. They are told that: Linda is 31 , single, outspoken, and very bright. She majored in philosophy in college. As a student, she was deeply concerned with racial discrimination and other social issues, and participated in anti-nuclear demonstrations. The subjects are then asked to rank the likelihood of various alternatives, such as: (1) Linda is active in the feminist movement. (2) Linda is a bank teller. (3) Linda is a bank teller and active in the feminist movement. Tversky and Kahneman found that between 85 and 90 percent of the subjects rated alternative (1) most likely, but alternative (3) more likely than alternative (2). Is it? They call this phenomenon the conjunction fallacy, and note that it appears to be unaffected by prior training in probability or statistics. Is this phenomenon a fallacy? If so, why?

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