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Chapter 6: Problem 13

Use the axioms for probability and mathematical induction to prove that for all integers \(n \geq 2\), if \(A_{1}, A_{2}, A_{3}, \ldots, A_{n}\) are any mutually disjoint events in a sample space \(S\), then $$ P\left(A_{1} \cup A_{2} \cup A_{3} \cup \cdots \cup A_{n}\right)=\sum_{k=1}^{n} P\left(A_{k}\right) $$

Short Answer

Expert verified
Using mathematical induction and the axioms of probability, we can prove that for all integers n 鈮� 2 and mutually disjoint events \(A_1,\dots,A_n\), the probability of their union is equal to the sum of their probabilities, i.e., \(P(A_1 \cup A_2 \cup \cdots \cup A_n) = \sum_{k=1}^n P(A_k)\). We first show that this formula holds true for two events (n=2) and then, by assuming it works for n=k, we prove that it also holds for n=k+1. Therefore, by induction, the formula is valid for all integers n 鈮� 2 as required.

Step by step solution

01

Base Case (n = 2)

For the base case when n = 2, we need to show that the formula works for two mutually disjoint events A1 and A2. Since A1 and A2 are disjoint, we have: \[P(A_1 \cup A_2) = P(A_1) + P(A_2)\] This is true, as it is one of the axioms of probability which states that the probability of the union of two disjoint events is equal to the sum of their probabilities.
02

Inductive Hypothesis

Now, we assume the formula is true for some value k, and we will need to prove that it holds for k+1 as well. So, let's assume \[P(A_1 \cup A_2 \cup \ldots \cup A_k) = \sum_{i=1}^k P(A_i)\] where A1, A2, ..., Ak are mutually disjoint events in a sample space S.
03

Inductive Step (Proving for n = k + 1)

We will now prove that this formula holds for n = k + 1. We want to find the probability of the union of k + 1 mutually disjoint events: \(P(A_1 \cup A_2 \cup \ldots \cup A_{k+1})\). Now, we know that: \[P(A_1 \cup A_2 \cup \ldots \cup A_k \cup A_{k+1}) = P((A_1 \cup A_2 \cup \ldots \cup A_k) \cup A_{k+1})\] Since \(A_1 \cup A_2 \cup \ldots \cup A_k\) is mutually disjoint with \(A_{k+1}\), we can write by the first axiom of probability as the sum: \[P((A_1 \cup A_2 \cup \ldots \cup A_k) \cup A_{k+1}) = P(A_1 \cup A_2 \cup \ldots \cup A_k) + P(A_{k+1})\] Now, using our inductive hypothesis for \(P(A_1 \cup A_2 \cup \ldots \cup A_k)\), we get: \[P(A_1 \cup A_2 \cup \ldots \cup A_k) + P(A_{k+1}) = \sum_{i=1}^k P(A_i) + P(A_{k+1})\] Therefore, the formula holds for n = k+1: \[P(A_1 \cup A_2 \cup \ldots \cup A_{k+1}) = \sum_{i=1}^{k+1} P(A_i)\]
04

Conclusion

By mathematical induction, we have proven that for all integers n 鈮� 2, if \(A_1, A_2, A_3, \ldots, A_n\) are any mutually disjoint events in a sample space S, then \[P(A_1 \cup A_2 \cup A_3 \cup \ldots \cup A_n) = \sum_{k=1}^n P(A_k)\]

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Key Concepts

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

Mutually Disjoint Events
Mutually disjoint events, also known as mutually exclusive events, are events in probability theory that cannot occur simultaneously. If you toss a six-sided die, the result can be a three or a four, but not both at the same time.

Mathematically, two events, say A and B, are considered mutually disjoint if their intersection is empty, shown as A 鈭� B = 脴. This concept is pivotal when dealing with the probability of combined events because the probability of the union of mutually disjoint events is simply the sum of their individual probabilities:
(P(A 鈭� B) = P(A) + P(B) if A and B are disjoint.

In more complex scenarios with multiple events, say A1, A2, A3, ..., An, they are mutually disjoint if every pair of events is disjoint, which leads to a similar calculation for their union's probability.
Mathematical Induction
Mathematical induction is a powerful proof technique used to prove that a statement holds for all natural numbers. It consists of two main steps: the base case and the inductive step.

To use mathematical induction, first, we show that the statement is true for the initial number, usually 1 or 2, which is the base case. Then, we assume that the statement holds for some arbitrary number k (inductive hypothesis) and show that it must also hold for k+1 (inductive step). If these steps are successful, we can conclude that the statement is true for all natural numbers.

The inductive step often involves algebraic manipulation using the inductive hypothesis to extend the truth of the statement to the next integer. This technique is particularly common in probability theory to show that properties hold over a range of scenarios.
Sample Space
The sample space of a probability experiment is the set of all possible outcomes. It provides a foundation for calculating probabilities and is represented by the symbol S. If you roll a six-sided die, for instance, the sample space is S = {1, 2, 3, 4, 5, 6}, with each number representing a potential outcome.

All individual events pertaining to a probability experiment can be considered subsets of this sample space. The concept of the sample space is critical because it helps define probability measures and ensures that probabilities of all possible outcomes together sum up to one, as stipulated by the axioms of probability.
Axioms of Probability
The axioms of probability are the foundation of probability theory, setting the rules for assigning probabilities to events. There are three primary axioms:
  1. The probability of an event is a non-negative number: P(A) 鈮� 0 for every event A.
  2. The probability of the sample space is 1: P(S) = 1.
  3. If two events A and B are mutually disjoint, the probability of their union is the sum of their probabilities: P(A 鈭� B) = P(A) + P(B), provided A 鈭� B = 脴.
These axioms help ensure consistency within probability and are utilized to derive more complex formulae, such as the one used in our exercise with mutually disjoint events across a sample space.

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Most popular questions from this chapter

a. How many positive two-digit integers are multiples of 3 ? b. What is the probability that a randomly chosen positive two-digit integer is a multiple of 3 ?

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