Q. 3.6 Prove that if E1,E2,鈥�,En聽are ... [FREE SOLUTION]

91影视

A First Course in Probability

Sheldon M. Ross

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 432 pages

ISBN: 9780321794772

Chapter 3: Q. 3.6 (page 107)

Prove that if $E_{1}, E_{2}, 鈥�, E_{n}$ are independent events, then
$P (E_{1} 鈭� E_{2} 鈭� 鈰� 鈭� E_{n}) = 1 - {鈭�}_{i = 1}^{n} [1 - P (E_{i})]$

Short Answer

Expert verified

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

$P (E_{1} 鈭� E_{2} 鈭� 鈰� 鈭� E_{n}) = 1 - {鈭�}_{i = 1}^{n} [1 - P (E_{i})]$ .

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 $E_{i}$

$P [{鈭�}_{i = 1}^{n} E_{i}] = 1 - {鈭�}_{i = 1}^{n} [1 - P (E_{i})]$

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

$P [{鈭�}_{i = 1}^{n} E_{i}] = {鈭�}_{i = 1}^{n} P (E_{i}) - {鈭�}_{i, j = 1 i < j}^{n} P (E_{i} E_{j}) + 鈥� + (- 1)^{n - 1} P (E_{1} E_{2} 鈥� E_{n})$

$= {鈭�}_{i = 1}^{n} P (E_{i}) - {鈭�}_{i, j = 1 i < j}^{n} P (E_{i}) P (E_{j}) + 鈥� + (- 1)^{n - 1} P (E_{1}) P (E_{2}) 鈥� P (E_{n})$

Final Answer

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

$1 - {鈭�}_{i = 1}^{n} [1 - P (E_{i})] = 1 - [1 - {鈭�}_{i = 1}^{n} P (E_{i}) + {鈭�}_{i, j = 1 i < j}^{n} P (E_{i}) P (E_{j}) + 鈥� + (- 1)^{n} P (E_{1}) P (E_{2}) 鈥� P (E_{n})$

$= {鈭�}_{i = 1}^{n} P (E_{i}) - {鈭�}_{i, j = 1 i < j}^{n} P (E_{i}) P (E_{j}) + 鈥� + (- 1)^{n - 1} P (E_{1} E_{2} 鈥� E_{n})$

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

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91影视

Prove that if $E_{1}, E_{2}, 鈥�, E_{n}$ are independent events, then
$P (E_{1} 鈭� E_{2} 鈭� 鈰� 鈭� E_{n}) = 1 - {鈭�}_{i = 1}^{n} [1 - P (E_{i})]$

Short Answer

Step by step solution

Concept Introduction

Explanation

Final Answer

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91影视

Prove that if E1,E2,鈥�,Enare independent events, thenPE1鈭�E2鈭�鈰�鈭�En=1-鈭�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

Geometry

Decision Maths

Probability and Statistics

Mechanics Maths

Applied Mathematics

Statistics

Study anywhere. Anytime. Across all devices.

Prove that if $E_{1}, E_{2}, 鈥�, E_{n}$ are independent events, then
$P (E_{1} 鈭� E_{2} 鈭� 鈰� 鈭� E_{n}) = 1 - {鈭�}_{i = 1}^{n} [1 - P (E_{i})]$