/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Show that $$ P\left(\bigcup_... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that $$ P\left(\bigcup_{i=1}^{n} E_{i}\right) \leqslant \sum_{i=1}^{n} P\left(E_{i}\right) $$ This is lenown as Boole's inequality. Hint: Either use Equation (1.2) and mathematical induction, or else show that \(\bigcup_{i=1}^{n} E_{l}=\bigcup_{i=1}^{n} F_{l}\), where \(F_{1}=E_{1}, F_{i}=E_{l} \bigcap_{j=1}^{i-1} E_{j}^{c}\), and use property (iii) of a probability.

Short Answer

Expert verified
To prove Boole's inequality, define a new set of events \(F_1, F_2, \dots, F_n\) as follows: $$F_1 = E_1$$ $$F_i = E_i \cap \bigcap_{j=1}^{i-1}E_j^c \text{, for } i = 2,3,\dots,n$$ Now, show that \(\bigcup_{i=1}^n E_i = \bigcup_{i=1}^n F_i\). Since the \(F_i\) sets are pairwise disjoint, we can use property (iii) of a probability: $$P\left(\bigcup_{i=1}^n F_i\right) = \sum_{i=1}^n P(F_i)$$ Finally, as \(P(F_i) \leq P(E_i)\), we can conclude: $$P\left(\bigcup_{i=1}^n E_i\right) \leq \sum_{i=1}^n P(E_i)$$ Hence, Boole's inequality is proved.

Step by step solution

01

Define the new sets \(F_i\)

First, let's define a new set of events \(F_1, F_2, \dots, F_n\) in terms of the given sets \(E_1, E_2, \dots, E_n\): $$F_1 = E_1$$ $$F_i = E_i \cap \bigcap_{j=1}^{i-1}E_j^c \text{, for } i = 2, 3,..., n$$ These new sets are such that \(F_i\) is the part of \(E_i\) that is not included in the union of the previous \(E_j\) events.
02

Express the union of \(E_i\) events in terms of \(F_i\) events

Now we want to show that $$\bigcup_{i=1}^n E_i = \bigcup_{i=1}^n F_i$$ Notice that since \(F_i\) is the part of \(E_i\) that is not included in \(\bigcup_{j=1}^{i-1} E_j\), the union of all \(E_i\) events will be the same as the union of all \(F_i\) events.
03

Apply property (iii)

Property (iii) of a probability states that if a set of events \(A_1, A_2, \dots, A_n\) are pairwise disjoint, then the probability of their union is equal to the sum of their individual probabilities: $$P\left(\bigcup_{i=1}^n A_i\right) = \sum_{i=1}^n P(A_i)$$ Since the sets \(F_i\) are pairwise disjoint (i.e., they do not have any common elements), we can use property (iii): $$P\left(\bigcup_{i=1}^n F_i\right) = \sum_{i=1}^n P(F_i)$$
04

Prove Boole's inequality

Since we know that \(\bigcup_{i=1}^n E_i = \bigcup_{i=1}^n F_i\), we also have $$P\left(\bigcup_{i=1}^n E_i\right) = P\left(\bigcup_{i=1}^n F_i\right)$$ Then, we can use the result from Step 3: $$P\left(\bigcup_{i=1}^n E_i\right) = \sum_{i=1}^n P(F_i)$$ Since each \(P(F_i) \leq P(E_i)\), this implies $$P\left(\bigcup_{i=1}^n E_i\right) = \sum_{i=1}^n P(F_i) \leq \sum_{i=1}^n P(E_i)$$ Thus, we have proved Boole's inequality.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Set Theory
Set theory provides a foundational framework for mathematics, especially in understanding collections of objects, referred to as "sets." Imagine sets as bags with certain items, and these items can be anything like numbers, functions, or so on. In this particular exercise, we deal with sets of events, such as \(E_1, E_2, \dots, E_n\). These events can represent possible outcomes in a probability scenario.

In set operations, the concept of the "union" of sets is crucial. The union \(\bigcup_{i=1}^{n} E_i\) represents a set containing all elements found in any \(E_i\). Conversely, when a set is complemented, like \(E_j^c\), it consists of all the elements not in \(E_j\). These operations form the backbone of expressing and manipulating events in probability and are instrumental in establishing the relationships between different events.
Probability Inequality
Probability inequalities serve as critical tools in measuring and comparing the likelihood of events. Boole's Inequality is a fundamental concept in probability that allows us to estimate the probability of the union of multiple events.

The inequality is expressed as:
  • \( P\left(\bigcup_{i=1}^{n} E_i\right) \leq \sum_{i=1}^{n} P(E_i) \)
This tells us that the probability of at least one of many events occurring is not greater than the sum of their individual probabilities.

An important aspect of this inequality is how it underscores the notion of independent and overlapping events. In scenarios where events might overlap (e.g., where one event being true doesn't make another event necessarily false), this inequality holds true. It acts as a safeguard to ensure our probability estimates don't exceed the logical limit of 1, which signifies certainty.
Mathematical Induction
Mathematical induction is a logical method used to prove statements or formulas that hold true for all natural numbers. Consider it like a domino effect: if you can prove a statement for one case (often the first) and then show that if it holds for one number \(k\), it will also hold for \(k+1\), you can conclude it's true for all subsequent numbers.

In the proof of Boole's Inequality, mathematical induction can be employed to show the inequality holds for any natural number \(n\). You begin by proving the base case, often \(n=1\), which is usually straightforward. Once established, you assume it's true for \(n=k\) and prove it must then be true for \(n=k+1\). Successfully completing these steps confirms the inequality's validity for all \(n\).

This methodical approach isn't just limited to numbers. It's a widely applicable technique in proving properties across mathematics, including probabilities, sets, and beyond.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events \(E_{1}, E_{2}, E_{3}\), and \(E_{4}\) as follows: $$ \begin{aligned} &E_{1}=\\{\text { the first pile has exactly } 1 \text { ace }\\} \\ &E_{2}=\\{\text { the second pile has exactly } 1 \mathrm{ace}\\} \\ &E_{3}=\\{\text { the third pile has exactly } 1 \text { ace }\\} \end{aligned} $$ \(E_{4}=\\{\) the fourth pile has exactly 1 ace\\}

Two cards are randomly selected from a deck of 52 playing cards. (a) What is the probability they constitute a pair (that is, that they are of the same denomination)? (b) What is the conditional probability they constitute a pair given that they are of different suits?

Um 1 contains two white balls and one black ball, while urn 2 contains one white ball and five black balls. One ball is drawn at random from urn 1 and placed in urn 2. A ball is then drawn from urn 2. It happens to be white. What is the probability that the transferred ball was white?

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. Now what is the probability that it is the fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now what is the probability that it is the fair coin?

Consider two boxes, one containing one black and one white marble, the other, two black and one white marble. A box is selected at random and a marble is drawn at random from the selected box. What is the probability that the marble is black?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.