Question

Suppose that A, B, and D are events such that A and B are independent,\({\bf{Pr}}\left( {{\bf{A}} \cap {\bf{B}} \cap {\bf{C}}} \right){\bf{ = 0}}{\bf{.04}}\),\({\bf{Pr}}\left( {{\bf{D|A}} \cap {\bf{B}}} \right){\bf{ = 0}}{\bf{.25}}\), and\({\bf{Pr}}\left( {\bf{B}} \right){\bf{ = 4Pr}}\left( {\bf{A}} \right)\). Evaluate\({\bf{Pr}}\left( {{\bf{A}} \cup {\bf{B}}} \right)\).

Short answer

The value of probability \(A \cup B\) is 0.34

Step-by-step solution

Given information

A,B and C are the three events and A and B are independent.

\(\Pr \left( {A \cap B \cap C} \right) = 0.04\), \(\Pr \left( {D|A \cap B} \right) = 0.25\) and \(\Pr \left( B \right) = 4\Pr \left( A \right)\)

Calculation for the required probability

\(\Pr \left( {A \cap B \cap C} \right) = 0.04\)

\(\Pr \left( {D|A \cap B} \right) = 0.25\)

\(\Pr \left( B \right) = 4\Pr \left( A \right)\)

So,

\(\begin{aligned}{}\Pr \left( {D|A \cap B} \right) = \frac{{\Pr \left( {D \cap A \cap B} \right)}}{{\Pr \left( {A \cap B} \right)}}\\ \Rightarrow \Pr \left( {A \cap B} \right) = \frac{{\Pr \left( {D \cap A \cap B} \right)}}{{\Pr \left( {D|A \cap B} \right)}}\\ \Rightarrow \Pr \left( {A \cap B} \right) = \frac{{0.04}}{{0.25}}\\ \Rightarrow \Pr \left( {A \cap B} \right) = 0.16\end{aligned}\)

Since we know that the event A and the event B are independent.

So,

\(\begin{aligned}{}\Pr \left( {A \cap B} \right) = \Pr \left( A \right)\Pr \left( B \right)\\0.16 = \Pr \left( A \right) \times 4\Pr \left( A \right)\\4\Pr {\left( A \right)^2} = 0.16\\\Pr \left( A \right) = \sqrt {\frac{{0.16}}{4}} \\\Pr \left( A \right) = 0.1\end{aligned}\)

So,

\(\begin{aligned}{}\Pr \left( {A \cup B} \right) = \Pr \left( A \right) + \Pr \left( B \right) - \Pr \left( {A \cap B} \right)\\ = \Pr \left( A \right) + 4\Pr \left( A \right) - 0.16\\ = 5\Pr \left( A \right) - 0.16\\ = \left( {5 \times 0.1} \right) - 0.16\\ = 0.5 - 0.16\\ = 0.34\end{aligned}\)

Therefore,\(\Pr \left( {A \cup B} \right) = 0.34\)