Chapter 3: Q. 3.6 (page 107)

Prove that if $E_{1}, E_{2}, \dots, E_{n}$ are independent events, then
$P (E_{1} \cup E_{2} \cup \dots \cup E_{n}) = 1 - \prod_{i = 1}^{n} [1 - P (E_{i})]$

Short Answer

Expert verified

By applying exclusion and inclusion we can prove that if $E_{1}, E_{2}, \dots, E_{n}$ are independent events then,

$P (E_{1} \cup E_{2} \cup \dots \cup E_{n}) = 1 - \prod_{i = 1}^{n} [1 - P (E_{i})]$ .

Step by step solution

Concept Introduction

Two possibilities are independent if the happening of one event does not affect the probabilities of the occurrence of the other event.

Explanation

Prove for independent $E_{i}$

$P [\cup_{i = 1}^{n} E_{i}] = 1 - \prod_{i = 1}^{n} [1 - P (E_{i})]$

Apply law of inclusion and exclusion, and independence on the left-hand side:

$P [\cup_{i = 1}^{n} E_{i}] = \sum_{i = 1}^{n} P (E_{i}) - \sum_{i, j = 1 i < j}^{n} P (E_{i} E_{j}) + \dots + (- 1)^{n - 1} P (E_{1} E_{2} \dots E_{n})$

$= \sum_{i = 1}^{n} P (E_{i}) - \sum_{i, j = 1 i < j}^{n} P (E_{i}) P (E_{j}) + \dots + (- 1)^{n - 1} P (E_{1}) P (E_{2}) \dots P (E_{n})$

Final Answer

On the right-hand side, by multiplying and then grouping by the number of $1$ 's in the product

$1 - \prod_{i = 1}^{n} [1 - P (E_{i})] = 1 - [1 - \sum_{i = 1}^{n} P (E_{i}) + \sum_{i, j = 1 i < j}^{n} P (E_{i}) P (E_{j}) + \dots + (- 1)^{n} P (E_{1}) P (E_{2}) \dots P (E_{n})$

$= \sum_{i = 1}^{n} P (E_{i}) - \sum_{i, j = 1 i < j}^{n} P (E_{i}) P (E_{j}) + \dots + (- 1)^{n - 1} P (E_{1} E_{2} \dots E_{n})$

As both sides of the starting equality are equal to the same number, this equality is proven.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Prove that if $E_{1}, E_{2}, \dots, E_{n}$ are independent events, then
$P (E_{1} \cup E_{2} \cup \dots \cup E_{n}) = 1 - \prod_{i = 1}^{n} [1 - P (E_{i})]$

Short Answer

Step by step solution

Concept Introduction

Explanation

Final Answer

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Mechanics Maths

Pure Maths

Theoretical and Mathematical Physics

Applied Mathematics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Prove that if E1,E2,…,Enare independent events, thenPE1∪E2∪⋯∪En=1-∏i=1n1-PEi

Short Answer

Step by step solution

Concept Introduction

Explanation

Final Answer

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Mechanics Maths

Pure Maths

Theoretical and Mathematical Physics

Applied Mathematics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Prove that if $E_{1}, E_{2}, \dots, E_{n}$ are independent events, then
$P (E_{1} \cup E_{2} \cup \dots \cup E_{n}) = 1 - \prod_{i = 1}^{n} [1 - P (E_{i})]$