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]