91Ó°ÊÓ

Events A and B are independent.

For two independent events A and B, the following relation holds:

\(\Pr \left( {A \cap B} \right) = \Pr \left( A \right) \cdot \Pr \left( B \right)\) - (1)

Thus, for eventsA^c and B^c to be independent, the following relation must be satisfied:

\(\Pr \left( {{A^C} \cap {B^C}} \right) = \Pr \left( {{A^C}} \right) \cdot \Pr \left( {{B^C}} \right)\)

Now, proving the above relation as:

\(\begin{aligned}{c}\Pr \left( {{A^C} \cap {B^C}} \right) &= \Pr {\left( {A \cup B} \right)^C}\\ &= 1 - \Pr \left( {A \cup B} \right)\\ &= 1 - \left\{ {\Pr \left( A \right) + \Pr \left( B \right) - \Pr \left( {A \cap B} \right)} \right\}\\ &= 1 - \Pr \left( A \right) - \Pr \left( B \right) + \Pr \left( {A \cap B} \right)\end{aligned}\)

Using equation (1) in the above expression as

\(\begin{aligned}{c}\Pr \left( {{A^C} \cap {B^C}} \right) &= 1 - \Pr \left( A \right) - \Pr \left( B \right) + \Pr \left( A \right) \cdot \Pr \left( B \right)\\ &= \left\{ {1 - \Pr \left( A \right)} \right\} - \Pr \left( B \right)\left\{ {1 - \Pr \left( A \right)} \right\}\\ &= \left\{ {1 - \Pr \left( A \right)} \right\}\left\{ {1 - \Pr \left( B \right)} \right\}\\ &= \Pr \left( {{A^C}} \right) \cdot \Pr \left( {{B^C}} \right)\end{aligned}\)

It shows that eventsA^c and B^c are also independent.