Question

Suppose that A and B are independent events such that\({\bf{Pr}}\left( {\bf{A}} \right){\bf{ = }}{\raise0.7ex\hbox{\({\bf{1}}\)} \!\mathord{\left/ {\vphantom {{\bf{1}} {\bf{3}}}}\right.}\!\lower0.7ex\hbox{\({\bf{3}}\)}}\)and\({\bf{Pr}}\left( {\bf{B}} \right){\bf{ > 0}}\). What is the value of\({\bf{Pr}}\left( {{\bf{A}} \cup {{\bf{B}}{\bf{c}}}{\bf{|B}}} \right)\)?

Short answer

The probability of \(A \cup {B^c}|B\) is 0.

Step-by-step solution

Given information

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 \(\Pr \left( B \right) > 0\)

Finding the value of \({\bf{Pr}}\left( {{\bf{A}} \cup {{\bf{B}}^{\bf{c}}}{\bf{|B}}} \right)\)

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 \(\Pr \left( {A \cup {B^c}|B} \right) = 0\).