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]