Show that \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(A' \cap B)\).

(i)     show that \({\text{P}}(A) = \frac{1}{3}\);

(ii)     hence find \({\text{P}}(B)\).

METHOD 1

\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)\)    M1

\( = {\text{P}}(A) + {\text{P}}(A \cap B) + {\text{P}}(A' \cap B) - {\text{P}}(A \cap B)\)    M1A1

\( = {\text{P}}(A) + {\text{P}}(A' \cap B)\)    AG

METHOD 2

\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A \cap B)\)    M1

\( = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A|B) \times {\text{P}}(B)\)    M1

\( = {\text{P}}(A) + \left( {1 - {\text{P}}(A|B)} \right) \times {\text{P}}(B)\)

\( = {\text{P}}(A) + {\text{P}}(A'|B) \times {\text{P}}(B)\)    A1

\( = {\text{P}}(A) + {\text{P}}(A' \cap B)\)    AG

[3 marks]

(i)     use \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(A' \cap B)\) and \({\text{P}}(A' \cap B) = {\text{P}}(B|A'){\text{P}}(A')\)     (M1)

\(\frac{4}{9} = {\text{P}}(A) + \frac{1}{6}\left( {1 - {\text{P}}(A)} \right)\)    A1

\(8 = 18{\text{P}}(A) + 3\left( {1 - {\text{P}}(A)} \right)\)    M1

\({\text{P}}(A) = \frac{1}{3}\)    AG

(ii)     METHOD 1

\({\text{P}}(B) = {\text{P}}(A \cap B) + {\text{P}}(A' \cap B)\)    M1

\( = {\text{P}}(B|A){\text{P}}(A) + {\text{P}}(B|A'){\text{P}}(A')\)    M1

\( = \frac{1}{3} \times \frac{1}{3} + \frac{1}{6} \times \frac{2}{3} = \frac{2}{9}\)    A1

METHOD 2

\({\text{P}}(A \cap B) = {\text{P}}(B|A){\text{P}}(A) \Rightarrow {\text{P}}(A \cap B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\)    M1

\({\text{P}}(B) = {\text{P}}(A \cup B) + {\text{P}}(A \cap B) - {\text{P}}(A)\)    M1

\({\text{P}}(B) = \frac{4}{9} + \frac{1}{9} - \frac{1}{3} = \frac{2}{9}\)    A1

[6 marks]