Question
Let $A_{1},{\text{ }A}_{2},\ldots,{\text{ }A}_{n}$ be independent events of the same sample space, then $P\left( A_{1} \cup {\text{ }A}_{2} \cup {\text{ }A}_{3} \cup \ldots \cup {\text{ }A}_{n} \right)$ is equal to
Options
Solution
Sol. [Key Concept: $P(A_{1} \cup \ldots \cup \ A_{n}) = 1 - \prod_{i}(1 - P(A_{i}))$.]
$$P\left( A_{1} \cup {\text{ }A}_{2} \cup {\text{ }A}_{3} \cup \ldots\ldots \cup {\text{ }A}_{n} \right) = 1 - P\left( A_{1} \cap {\text{ }A}_{2} \cap {\text{ }A}_{3} \cap \ldots\ldots \cap {\text{ }A}_{n} \right)$$
$$\begin{matrix} & = 1 - P\left( {\text{ }A}_{1} \right)P\left( {\text{ }A}_{2} \right)\ldots..P\left( {\text{ }A}_{n} \right) \\ & \ = 1 - \left( \left( 1 - P\left( {\text{ }A}_{1} \right) \right)\left( 1 - P\left( {\text{ }A}_{2} \right) \right)\ldots\ldots.\left( 1 - P\left( {\text{ }A}_{n} \right) \right) \right.\ \end{matrix}$$
Create a free account to view solution
View Solution Free