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Expert-verified Found in: Page 467 ### Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095 # Question: Use mathematical induction to prove that ${{\mathbit{E}}}_{{\mathbf{1}}}{\mathbf{,}}{{\mathbit{E}}}_{{\mathbf{2}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{{\mathbit{E}}}_{{\mathbf{n}}}$ is a sequence of ${\mathbit{n}}$ pair wise disjoint events in a sample space ${\mathbit{S}}$, where ${\mathbit{n}}$ is a positive integer, then ${\mathbf{}}{\mathbit{p}}\left({\cup }_{i=1}^{n}{E}_{i}\right){\mathbf{=}}{{\mathbf{\sum }}}_{\mathbf{i}\mathbf{=}\mathbf{1}}^{{\mathbf{n}}}{\mathbit{p}}\left({E}_{i}\right)$ .

It is proved that $p\left({\cup }_{i=1}^{n}{E}_{i}\right)=\sum _{i=1}^{n}p\left({E}_{i}\right)$.

See the step by step solution

## 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: ${\mathbit{p}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ is true

Inductive Step: if ${\mathbit{p}}\mathbf{\left(}\mathbf{n}\mathbf{\right)}$ is true, then ${\mathbit{p}}\left(n+1\right)$ is true

Then, ${\mathbit{p}}\mathbf{\left(}\mathbf{n}\mathbf{\right)}$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 $n=1$

$p\left({E}_{1}\right)=p\left({E}_{1}\right)$

For $n=2$

$p\left({E}_{1}{\cup E}_{2}\right)=p\left({E}_{1}\right)+p\left({E}_{2}\right)-p\left({E}_{1}{\cap E}_{2}\right)$

As ${E}_{1},{E}_{2}$ are disjoint events ${E}_{1}{\cap E}_{2}=\varphi$

$p\left({E}_{1}{\cap E}_{2}\right)=0$

Applying the result in equation

$P\left({E}_{1}{\cup E}_{2}\right)=P\left({E}_{1}\right)+P\left({E}_{2}\right)$

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

Assume it is true for $n=k$

$p\left({E}_{1}{\cup E}_{2}{\cup E}_{3}...{\cup E}_{K}\right)=p\left({E}_{1}\right)+p\left({E}_{2}\right)+p\left({E}_{3}\right)+...+p\left({E}_{K}\right)$

Prove the result for $n=k+1$

$p\left({E}_{1}{\cup E}_{2}{\cup E}_{3}...{\cup E}_{K+1}\right)=p\left({E}_{1}\right)+p\left({E}_{2}\right)+p\left({E}_{3}\right)+...+p\left({E}_{K+1}\right)\phantom{\rule{0ex}{0ex}}p\left({E}_{1}{\cup E}_{2}{\cup E}_{3}...{\cup E}_{K}{\cup E}_{K+1}\right)=p\left(\left({E}_{1}{\cup E}_{2}{\cup E}_{3}...{\cup E}_{K}\right)\cup \left({E}_{K+1}\right)\right)\phantom{\rule{0ex}{0ex}}$

Use equation $\left(2\right)$ and $\left(3\right)$ in the above equation

$p\left({E}_{1}{\cup E}_{2}{\cup E}_{3}...{\cup E}_{K+1}\right)=p\left({E}_{1}\right)+p\left({E}_{2}\right)+p\left({E}_{3}\right)+...+p\left({E}_{K}\right)+p\left({E}_{K+1}\right)$

It is true for $n=k+1$

So, $p\left({\cup }_{i=1}^{n}{E}_{i}\right)=\sum _{i=1}^{n}p\left({E}_{i}\right)$

Here $n$ is a positive integer

Thus, it is proved that $p\left({\cup }_{i=1}^{n}{E}_{i}\right)=\sum _{i=1}^{n}p\left({E}_{i}\right)$ . ### Want to see more solutions like these? 