Write down an expression for \({\text{P}}(A \cap B)\) .

Given that \({\text{P}}(A \cup B) = 0.8\) ,

(i)     find x ;

(ii)    find \({\text{P}}(A \cap B)\) .

Hence, explain why A and B are not mutually exclusive.

\({\text{P}}(A \cap B) = {\text{P}}(A) \times {\text{P}}(B)( = 0.6x)\)    A1     N1

[1 mark]

(i) evidence of using \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) - {\text{P}}(A){\text{P}}(B)\)     (M1)

correct substitution     A1

e.g. \(0.8 = 0.6 + x - 0.6x\) , \(0.2 = 0.4x\)

\(x = 0.5\)    A1     N2

(ii) \({\text{P}}(A \cap B) = 0.3\)     A1     N1

[4 marks]

valid reason, with reference to \({\text{P(}}A \cap B)\)     R1     N1

e.g. \({\text{P(}}A \cap B) \ne 0\)

[1 mark]

This question was well done by most candidates.

This question was well done by most candidates. When errors were made, candidates confused the terms "independent" and "mutually exclusive" and did not subtract the intersection when finding \({\text{P}}(A \cup B)\) .

Candidates should also be aware of the command term "hence" used in part (c) where they were expected to provide a reason that involved \({\text{P}}(A \cap B)\) from their work in part (b). It seemed that many turned to the formula in the booklet instead of considering the conceptual meaning of the term.