91Ó°ÊÓ

A and B are the two independent events.

\(\Pr \left( A \right) = {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 3}}\right.}\!\lower0.7ex\hbox{$3$}}\) and \(\Pr \left( B \right) > 0\)

Let,\(C = \left( {A \cup {B^c}} \right)\)

So,

\(\Pr \left( {A \cup {B^c}|B} \right) = \frac{{\Pr \left( {\left( {A \cup {B^c}} \right) \cap B} \right)}}{{\Pr \left( B \right)}}\)

i.e.,\(\Pr \left( {A \cup {B^c}|B} \right) = \frac{{\Pr \left( {C \cap B} \right)}}{{\Pr \left( B \right)}}\)

Here, the event B and the event C are the disjoint events.

So,\(\Pr \left( {C \cap B} \right) = 0\)

\(\begin{aligned}{}\Pr \left( {A \cup {B^c}|B} \right) = \frac{0}{{\Pr \left( B \right)}}\\ = 0\end{aligned}\)

Since we know that,\(\Pr \left( B \right) > 0\)

Therefore, required probability is \(\Pr \left( {A \cup {B^c}|B} \right) = 0\).