Book edition 7th

Author(s) Kenneth H. Rosen

Pages 808 pages

ISBN 9780073383095

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Short Answer

Question: Use mathematical induction to prove that $E_{1}, E_{2}, . . ., E_{n}$ is a sequence of $n$ pair wise disjoint events in a sample space $S$ , where $n$ is a positive integer, then $p (\cup_{i = 1}^{n} E_{i}) = \sum_{i = 1}^{n} p (E_{i})$ .

Answer:

It is proved that $p (\cup_{i = 1}^{n} E_{i}) = \sum_{i = 1}^{n} p (E_{i})$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

It is given that that $E_{1}, E_{2}, . . ., E_{n}$ is a sequence of $n$ pair wise disjoint events in a sample space $S$ , here $n$ is a positive integer.

Step 2: Definition and formula to be used

Principle of mathematical Induction states that “Let be a property of positive integers such that,

Basic Step: $p (1)$ is true

Inductive Step: if $p (n)$ is true, then $p (n + 1)$ is true

Then, $p (n)$ is true for all positive integers.”

Step 3: Apply the principle of mathematical induction

By using mathematical induction prove the result for $n = 1$

For $n = 1$

$p (E_{1}) = p (E_{1})$

For $n = 2$

$p (E_{1} \cupE_{2}) = p (E_{1}) + p (E_{2}) - p (E_{1} \capE_{2})$

As $E_{1}, E_{2}$ are disjoint events $E_{1} \capE_{2} = ϕ$

$p (E_{1} \capE_{2}) = 0$

Applying the result in equation

$P (E_{1} \cupE_{2}) = P (E_{1}) + P (E_{2})$

Thus, the result is true for $n = 2$

Assume it is true for $n = k$

$p (E_{1} \cupE_{2} \cupE_{3} . . . \cupE_{K}) = p (E_{1}) + p (E_{2}) + p (E_{3}) + . . . + p (E_{K})$

Prove the result for $n = k + 1$

$p (E_{1} \cupE_{2} \cupE_{3} . . . \cupE_{K + 1}) = p (E_{1}) + p (E_{2}) + p (E_{3}) + . . . + p (E_{K + 1}) p (E_{1} \cupE_{2} \cupE_{3} . . . \cupE_{K} \cupE_{K + 1}) = p ((E_{1} \cupE_{2} \cupE_{3} . . . \cupE_{K}) \cup (E_{K + 1}))$

Use equation $(2)$ and $(3)$ in the above equation

$p (E_{1} \cupE_{2} \cupE_{3} . . . \cupE_{K + 1}) = p (E_{1}) + p (E_{2}) + p (E_{3}) + . . . + p (E_{K}) + p (E_{K + 1})$

It is true for $n = k + 1$

So, $p (\cup_{i = 1}^{n} E_{i}) = \sum_{i = 1}^{n} p (E_{i})$

Here $n$ is a positive integer

Thus, it is proved that $p (\cup_{i = 1}^{n} E_{i}) = \sum_{i = 1}^{n} p (E_{i})$ .

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Discrete Mathematics and its Applications

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Short Answer

Question: Use mathematical induction to prove that $E_{1}, E_{2}, . . ., E_{n}$ is a sequence of $n$ pair wise disjoint events in a sample space $S$ , where $n$ is a positive integer, then $p (\cup_{i = 1}^{n} E_{i}) = \sum_{i = 1}^{n} p (E_{i})$ .

Step by Step Solution

Step 1: Given Information

Step 2: Definition and formula to be used

Step 3: Apply the principle of mathematical induction

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Step by Step Solution

Step 1: Given Information

Step 2: Definition and formula to be used

Step 3: Apply the principle of mathematical induction

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Question: Use mathematical induction to prove that $E_{1}, E_{2}, . . ., E_{n}$ is a sequence of $n$ pair wise disjoint events in a sample space $S$ , where $n$ is a positive integer, then $p (\cup_{i = 1}^{n} E_{i}) = \sum_{i = 1}^{n} p (E_{i})$ .