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Q36E

Expert-verifiedFound in: Page 467

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Question: Use mathematical induction to prove that ${{\mathit{E}}}_{{\mathbf{1}}}{\mathbf{,}}{{\mathit{E}}}_{{\mathbf{2}}}{\mathbf{,}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{,}}{{\mathit{E}}}_{{\mathbf{n}}}$ is a sequence of ${\mathit{n}}$ pair wise disjoint events in a sample space ${\mathit{S}}$, where ${\mathit{n}}$ is a positive integer, then ${\mathbf{}}{\mathit{p}}{\left({\cup}_{i=1}^{n}{E}_{i}\right)}{\mathbf{=}}{{\mathbf{\sum}}}_{\mathbf{i}\mathbf{=}\mathbf{1}}^{{\mathbf{n}}}{\mathit{p}}{\left({E}_{i}\right)}$ .**

**Answer:**

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

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.

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

Basic Step: ${\mathit{p}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ is true

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

Then, ${\mathit{p}}\mathbf{\left(}\mathbf{n}\mathbf{\right)}$is true for all positive integers.”

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

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