ProbabilityHardBloom L3

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

A.$\prod_{i = 1}^{n}\mspace{2mu}\left( 1 - P\left( A_{i} \right) \right.\ $
B.$1 - \prod_{i = 1}^{n}\mspace{2mu}\left( 1 - P\left( A_{i} \right) \right.\ $
C.$1 - P\left( \overline{A_{1} \cap A_{2} \cap A_{5} \cap \ldots\ldots \cap A_{n}} \right)$
D.$1 - P\left( {\overline{A}}_{1} \cap {\overline{\text{ }A}}_{2} \cap {\overline{\text{ }A}}_{3} \cap \ldots\ldots \cap {\overline{\text{ }A}}_{n} \right)$

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
Topic: Probability·Practice all Probability questions

More Probability Questions