Question

Prove that if $E_1,E_2,\dots ,E_n$are independent events, then

$P\left(E_1\cup E_2\cup \cdots \cup E_n\right)=1-\prod _{i=1}^n\left[1-P\left(E_i\right)\right]$

Short answer

By applying exclusion and inclusion we can prove that if $E_1,E_2,\dots ,E_n$are independent events then,

$P\left(E_1\cup E_2\cup \cdots \cup E_n\right)=1-\prod _{i=1}^n\left[1-P\left(E_i\right)\right]$.

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 $E_i$

$P\left[\underset{i=1}{\overset{n}{\cup }}{E}_i\right]=1-\prod _{i=1}^n\left[1-P\left(E_i\right)\right]$

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

$$\begin{gathered} P\left[\underset{i=1}{\overset{n}{\cup }}{E}_i\right]=\sum _{i=1}^{n}P\left(E_i\right)-\sum _{i,j=1 \\ i<j}^{n}P\left(E_i{E}_j\right)+\dots +(-1{)}^{n-1}P\left(E_1{E}_2\dots E_n\right) \end{gathered}$$

$$\begin{gathered} =\sum _{i=1}^{n}P\left(E_i\right)-\sum _{i,j=1 \\ i<j}^{n}P\left(E_i\right)P\left(E_j\right)+\dots +(-1{)}^{n-1}P\left(E_1\right)P\left(E_2\right)\dots P\left(E_n\right) \end{gathered}$$

Final Answer

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

$$\begin{gathered} 1-\prod _{i=1}^n\left[1-P\left(E_i\right)\right]=1-\left[1-\sum _{i=1}^{n}P\left(E_i\right)+\sum _{i,j=1 \\ i<j}^{n}P\left(E_i\right)P\left(E_j\right)+\dots +(-1{)}^{n}P\left(E_1\right)P\left(E_2\right)\dots P\left(E_n\right)\right. \end{gathered}$$

$$\begin{gathered} =\sum _{i=1}^{n}P\left(E_i\right)-\sum _{i,j=1 \\ i<j}^{n}P\left(E_i\right)P\left(E_j\right)+\dots +(-1{)}^{n-1}P\left(E_1{E}_2\dots E_n\right) \end{gathered}$$

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