91Ó°ÊÓ

Independence If the probability of one event is unaffected by the occurrence or non-occurrence of the other, two events are said to be independent in probability. Consider the following example for a better understanding of this definition.

Denote event \({\rm{A = \{ }}\) the division is incorrect \({\rm{\} }}\). We are given the probability of event \({\rm{A,P(A) = 0}}{\rm{.0000000009}}\) (there are \({\rm{9}}\) zeros) or

\({\rm{P(A) = }}\frac{{\rm{1}}}{{{\rm{9,000,000,000}}}}{\rm{ = a}}\)

one in nine billion.

Assume that the same chip has one billion divisions.

Denote events \({{\rm{A}}_{\rm{i}}}{\rm{ = }}\left\{ {} \right.\)the ith division is incorrect \(\} \) from \({\rm{i = 1}}\) to \({\rm{i = }}\)billion \(\left( {{\rm{i = 1,2, \ldots ,1}}{{\rm{0}}^{\rm{9}}}} \right)\).

With this chip, we must determine the likelihood of at least one error occurring in one billion divisions or the union of the billion \({{\rm{A}}_{\rm{i}}}\)events.

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{{{10}^9}}}} \right.\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P}}\left( {{{\left( {A_1^\prime \cap A_2^\prime \cap \ldots \cap A_{{{10}^9}}^\prime } \right)}^\prime }} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{1 - P}}\left( {A_1^\prime \cap A_2^\prime \cap {\rm{ \ldots }} \cap A_{{{10}^9}}^\prime } \right)\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\rm{1 - P}}\left( {A_1^\prime } \right)*P\left( {A_2^\prime } \right){\rm{* \ldots *P}}\left( {A_{{{10}^9}}^\prime } \right)\\\mathop {\rm{ = }}\limits^{{\rm{(4)}}} {\rm{1 - (1 - a)*(1 - a)* \ldots *(1 - a)}}\\{\rm{ = 1 - 0}}{\rm{.895}}\\{\rm{ = 0}}{\rm{.105}}{\rm{.}}\end{array}\)

(1): De Morgan's Law is applied here.

(2): for any event \({\rm{A,P}}\left( {{A^\prime }} \right){\rm{ + P(A) = 1}}\),

(3): Because the events (points) are independent, we may utilise the following multiplication property.

(4): using \({\rm{A,P}}\left( {{A^\prime }} \right){\rm{ + P(A) = 1}}\).

Property of Multiplication:

For events \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{A}}_{\rm{n}}}{\rm{,n}} \in {\rm{N}}\) If they are mutually independent, we say they are mutually reliant

\(P\left( {{A_{{i_1}}} \cap {A_{{i_2}}} \ldots {A_{{i_k}}}} \right) = P\left( {{A_{{i_1}}}} \right) \cdot P\left( {{A_{{i_2}}}} \right) \cdot \ldots \cdot P\left( {{A_{{i_k}}}} \right)\)

for every \({\rm{k}} \in {\rm{\{ 2,3, \ldots ,n\} }}\), and every subset of indices \({{\rm{i}}_{\rm{1}}}{\rm{,}}{{\rm{i}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{i}}_{\rm{k}}}\).

\({\rm{P}}\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{{{10}^9}}}} \right){\rm{ = 0}}{\rm{.105}}.\)

Thus, the probability that at least one error occurs\({\rm{ = 0}}{\rm{.105}}\).