Question

Question 2. To show

If${\mathit{E}}_{\mathbf{1}}\mathbf{,}{\mathit{E}}_{\mathbf{2}}\mathbf{,}{\mathit{E}}_{\mathbf{3}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}{\mathit{E}}_{\mathbf{n}}$are events from a finite sample space, then$\mathbf{}\mathit{p}\mathbf{(}{\mathit{E}}_{\mathbf{1}}\mathbf{\cup }{\mathit{E}}_{\mathbf{2}}\mathbf{\cup }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\cup }{\mathit{E}}_{\mathbf{n}}\mathbf{)}\mathbf{⩽}\mathit{P}\mathbf{\left(}{\mathit{E}}_{\mathbf{1}}\mathbf{\right)}\mathbf{+}\mathit{p}\mathbf{\left(}{\mathit{E}}_{\mathbf{2}}\mathbf{\right)}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathit{p}\mathbf{\left(}{\mathit{E}}_{\mathbf{n}}\mathbf{\right)}\mathbf{.}$

Short answer

Answer

The expression is true$p(E_1\cup E_2\cup .....\cup E_n)⩽P\left(E_1\right)+p\left(E_2\right)+....+p\left(E_n\right).$

Step-by-step solution

Step 1. Given information

$E$are the Events from a finite sample space.

Step 2. Definition and formula to be used

According to Boole’s inequality,

$P(\cup [E_i\left]\right)⩽\left(P\right(E_i\left)\right)$

For n=1, we have,

$P\left(E_1\right)⩽P\left(E_1\right)$

This is true in all case.

Let us assume that the result holds true for n=k.

$P\left({\cup }_{(i=1)}^k\right[E_{i]}\left]\right)⩽{\sum }_{(i=1)}^k\left(P{E}_i\right)$

Step 3. Use Boole’s inequality

We will prove that the result is also true for$n=k+1$

Consider,

$P(X\cup Y)=P\left(X\right)+P\left(Y\right)-P(X\cap Y)$

So, we have,

$P\left({\cup }_{i=1}^{(k+1)}\right[E_i\left]\right)=P\left({\cup }_{(i=1)}^k\right[E_i\left]\right)+P\left(E_{(k+1)}\right)-P\left({\cup }_{(i=1)}^k\right[E_i]\cap (E_{(k+1)}$

We know that,

$P\left({\cup }_{(i=1)}^k\right[E_i\left]\right)\cap E_{(k+1)}⩾0$

Therefore, we have

$$\begin{gathered} P\left({\cup }_{i=1}^{(k+1)}\right[E_i\left]\right)⩽P\left({\cup }_{(i=1)}^k\right[E_i\left]\right)+P\left(E_{(k+1)}\right) \\ P\left({\cup }_{(i=1)}^{(k+1)}\right[E_i]⩽\sum _{(i=1)}^k\left(P\right(E_i\left)\right)+P\left(E_{k+1}\right) \\ P({\cup }_{i=1}^{k+1}\left[E_i\right])⩽{\sum }_{(i=1)}^{(k+1)}\left(P\right(E_i)) \end{gathered}$$

Since is result is proven to be true for,$n=k+1$

When it is true for$n=1$

and$n=k$, therefore, the results hold true for all$n$