Find ${\text{P}}(B)$.

Find ${\text{P}}(A)$.

Hence show that events $A’$ and $B$ are independent.

${\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}$

$\Rightarrow 0.75 = \frac{{0.6}}{{{\text{P}}(B)}}$     (M1)

$\Rightarrow {\text{P}}(B){\text{ }}\left( { = \frac{{0.6}}{{0.75}}} \right) = 0.8$     A1

[2 marks]

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

$\Rightarrow 0.95 = {\text{P}}(A) + 0.8 - 0.6$     (M1)

$\Rightarrow {\text{P}}(A) = 0.75$     A1

[2 marks]

METHOD 1

${\text{P}}(A'|B) = \frac{{{\text{P}}(A' \cap B)}}{{{\text{P}}(B)}} = \frac{{0.2}}{{0.8}} = 0.25$     A1

${\text{P}}(A'|B) = {\text{P}}(A’)$     R1

hence $A’$ and $B$ are independent     AG

Note:     If there is evidence that the student has calculated ${\text{P}}(A' \cap B) = 0.2$ by assuming independence in the first place, award A0R0.

METHOD 2

EITHER

${\text{P}}(A) = {\text{P}}(A|B)$     A1

OR

${\text{P}}(A) \times {\text{P}}(B) = 0.75 \times 0.80 = 0.6 = {\text{P}}(A \cap B)$     A1

THEN

$A$ and $B$ are independent     R1

hence $A’$ and $B$ are independent     AG

METHOD 3

${\text{P}}(A') \times {\text{P}}(B) = 0.25 \times 0.80 = 0.2$     A1

${\text{P}}(A') \times {\text{P}}(B) = {\text{P}}(A' \cap B)$     R1

hence $A’$ and $B$ are independent     AG

[2 marks]